Explanation:
Okay, so I need to find P(16) where P(x) is a quadratic polynomial that's squeezed between two other quadratics: x² - 2x + 2 and 2x² - 4x + 3 for all real numbers x. Also, they told me that P(11) = 181. Hmm, let me think about how to approach this.

First, since P(x) is a quadratic polynomial, I can write it in the general form: P(x) = ax² + bx + c, where a, b, c are real numbers. The problem says that for all real x, x² - 2x + 2 ≤ P(x) ≤ 2x² - 4x + 3. So, this inequality must hold for every x.

Since both the lower and upper bounds are quadratics, maybe P(x) is a quadratic that lies between them. But how can I find the coefficients a, b, c? Well, maybe I can consider the difference between P(x) and the lower bound, and the difference between the upper bound and P(x). Let me try that.

Let me denote the lower bound as L(x) = x² - 2x + 2 and the upper bound as U(x) = 2x² - 4x + 3. Then, for all x, L(x) ≤ P(x) ≤ U(x).

Since P(x) is a quadratic, maybe the difference between P(x) and L(x) is always non-negative, and the difference between U(x) and P(x) is also always non-negative. Let me write those differences:

P(x) - L(x) = (ax² + bx + c) - (x² - 2x + 2) = (a - 1)x² + (b + 2)x + (c - 2)

Similarly, U(x) - P(x) = (2x² - 4x + 3) - (ax² + bx + c) = (2 - a)x² + (-4 - b)x + (3 - c)

Since these differences must be non-negative for all real x, that means that each of these quadratics must be non-negative for all x. For a quadratic to be non-negative for all real x, it must satisfy two conditions: the leading coefficient is non-negative, and the discriminant is less than or equal to zero.

So, for P(x) - L(x) to be non-negative for all x:

1. The leading coefficient (a - 1) must be ≥ 0 ⇒ a ≥ 1
2. The discriminant must be ≤ 0. Let me compute the discriminant for P(x) - L(x):

Discriminant D1 = (b + 2)² - 4*(a - 1)*(c - 2) ≤ 0

Similarly, for U(x) - P(x) to be non-negative for all x:

1. The leading coefficient (2 - a) must be ≥ 0 ⇒ 2 - a ≥ 0 ⇒ a ≤ 2
2. The discriminant must be ≤ 0. Let me compute the discriminant for U(x) - P(x):

Discriminant D2 = (-4 - b)² - 4*(2 - a)*(3 - c) ≤ 0

So, combining the two inequalities from the leading coefficients, we have 1 ≤ a ≤ 2. Therefore, a is between 1 and 2.

Now, since P(x) is squeezed between L(x) and U(x) for all x, maybe P(x) is a linear combination of L(x) and U(x)? Or perhaps P(x) touches both bounds at some points? Wait, but since it's a quadratic, maybe it's tangent to both bounds? Or maybe it's equal to both bounds at some points?

Alternatively, since the inequalities hold for all x, maybe the difference between P(x) and L(x) is a non-negative quadratic, and the difference between U(x) and P(x) is also a non-negative quadratic. Therefore, both quadratics must be perfect squares? Or maybe they have a double root?

But maybe there's another approach. Let me think. If P(x) is between L(x) and U(x) for all x, then perhaps P(x) is equal to L(x) + something and U(x) - something else. But since both L(x) and U(x) are quadratics, maybe P(x) is a convex combination of them? Wait, but convex combinations would require coefficients that sum to 1, but here we don't know.

Alternatively, since both inequalities must hold for all x, maybe the minimum of U(x) - P(x) is zero and the maximum of P(x) - L(x) is zero? Wait, but that might not necessarily be the case. However, if P(x) is squeezed between L(x) and U(x) for all x, then maybe P(x) is equal to L(x) at some points and equal to U(x) at some other points? But since both L(x) and U(x) are quadratics, and P(x) is a quadratic, maybe they intersect at two points?

Alternatively, since the inequalities hold for all x, then the difference between P(x) and L(x) is non-negative for all x, and the difference between U(x) and P(x) is non-negative for all x. Therefore, both quadratics P(x) - L(x) and U(x) - P(x) are non-negative definite. Therefore, they must be perfect squares? Or at least have discriminant zero?

Wait, but if a quadratic is non-negative for all x, it can be written as k*(quadratic with leading coefficient positive) where the discriminant is zero? Or maybe not necessarily perfect squares. For example, if the discriminant is negative, then it's always positive. But if discriminant is zero, then it's a perfect square. However, since we have two different quadratics, maybe they have different discriminants?

But maybe there's a way to find P(x) by considering that the inequalities hold for all x. Let me think. Let me suppose that P(x) is equal to L(x) + k*(U(x) - L(x)) for some constant k between 0 and 1. Wait, that might be a way to express P(x) as a convex combination of L(x) and U(x). Let me check.

Let me compute U(x) - L(x):

U(x) - L(x) = (2x² - 4x + 3) - (x² - 2x + 2) = x² - 2x + 1 = (x - 1)^2.

Oh! That's interesting. So the difference between the upper and lower bounds is a perfect square. Therefore, U(x) - L(x) = (x - 1)^2. Therefore, if I write P(x) as L(x) + k*(x - 1)^2, where k is between 0 and 1, then P(x) would lie between L(x) and U(x). Because if k is 0, P(x) = L(x); if k is 1, P(x) = U(x). And for 0 < k < 1, P(x) is between them.

But since P(x) is a quadratic, and L(x) is quadratic, adding k*(x - 1)^2 would still give a quadratic. Therefore, maybe P(x) is of the form L(x) + k*(x - 1)^2. Let me check if this works.

Let me write that down:

P(x) = L(x) + k*(x - 1)^2 = x² - 2x + 2 + k*(x² - 2x + 1) = (1 + k)x² + (-2 - 2k)x + (2 + k)

Alternatively, since U(x) = L(x) + (x - 1)^2, so P(x) = L(x) + k*(x - 1)^2, where 0 ≤ k ≤ 1.

But then, since P(x) is a quadratic, and we have P(11) = 181, we can use that to find k. Let me try that.

So, let me compute P(11):

P(11) = (1 + k)*(11)^2 + (-2 - 2k)*(11) + (2 + k)

Let me compute each term:

First term: (1 + k)*121 = 121 + 121k

Second term: (-2 - 2k)*11 = -22 - 22k

Third term: 2 + k

Adding them all together:

121 + 121k - 22 - 22k + 2 + k = (121 - 22 + 2) + (121k - 22k + k) = 101 + 100k

So, P(11) = 101 + 100k. But we are told that P(11) = 181. Therefore:

101 + 100k = 181 ⇒ 100k = 80 ⇒ k = 80 / 100 = 0.8 = 4/5.

So, k = 4/5. Therefore, P(x) = L(x) + (4/5)*(x - 1)^2. Let me write that out:

P(x) = x² - 2x + 2 + (4/5)(x² - 2x + 1) = Let me compute that:

First, expand (4/5)(x² - 2x + 1) = (4/5)x² - (8/5)x + 4/5

Then add to L(x):

x² - 2x + 2 + (4/5)x² - (8/5)x + 4/5

Combine like terms:

x² + (4/5)x² = (9/5)x²

-2x - (8/5)x = (-10/5 - 8/5)x = (-18/5)x

2 + 4/5 = 10/5 + 4/5 = 14/5

Therefore, P(x) = (9/5)x² - (18/5)x + 14/5.

Alternatively, we can factor out 1/5:

P(x) = (1/5)(9x² - 18x + 14).

But let me check if this is correct. Let me verify if this P(x) satisfies the original inequalities.

First, check if P(x) is between L(x) and U(x) for all x. Since we constructed it as L(x) + k*(x - 1)^2 with 0 < k < 1, then yes, it should be between them. But let me check for a specific value. For example, at x = 1, L(1) = 1 - 2 + 2 = 1, U(1) = 2 - 4 + 3 = 1. So both bounds are equal to 1 at x = 1. Then P(1) = (9/5)(1) - (18/5)(1) + 14/5 = (9 - 18 + 14)/5 = 5/5 = 1. So P(1) = 1, which is equal to both bounds. That makes sense because the difference between U and L is (x - 1)^2, so at x = 1, the difference is zero, so P(x) must be equal to both bounds there.

Another check: Let me check x = 0. L(0) = 0 - 0 + 2 = 2. U(0) = 0 - 0 + 3 = 3. P(0) = 0 - 0 + 14/5 = 14/5 = 2.8, which is between 2 and 3. Good.

Another check: Let me check x = 2. L(2) = 4 - 4 + 2 = 2. U(2) = 8 - 8 + 3 = 3. P(2) = (9/5)(4) - (18/5)(2) + 14/5 = 36/5 - 36/5 + 14/5 = 14/5 = 2.8. Again between 2 and 3.

So seems okay. Now, since we found k = 4/5, then P(x) is determined. Therefore, we can compute P(16) using this expression.

So, let me compute P(16):

P(16) = (9/5)(16)^2 - (18/5)(16) + 14/5

First compute each term:

(9/5)(256) = (9 * 256)/5

(18/5)(16) = (18 * 16)/5

So:

P(16) = (9*256 - 18*16 + 14)/5

Let me compute numerator:

First, compute 9*256:

Well, 256 * 9: 256*10 = 2560, minus 256 = 2560 - 256 = 2304

Then 18*16 = 288

So numerator is 2304 - 288 + 14 = Let's compute 2304 - 288 first:

2304 - 288: 2304 - 200 = 2104; 2104 - 88 = 2016

Then 2016 + 14 = 2030

Therefore, P(16) = 2030 / 5 = 406.

Wait, 2030 divided by 5 is 406? Let me check:

5 * 400 = 2000, so 2030 - 2000 = 30, so 30 / 5 = 6. Therefore, 400 + 6 = 406. Yes, correct.

So P(16) = 406. Therefore, the answer is 406.

But let me check again if my expression for P(x) is correct. Let me verify with another point. For example, let me check P(11) = 181.

Earlier, we had P(11) = 101 + 100k, and with k = 4/5, that gives 101 + 80 = 181. Correct. So that's correct.

Alternatively, compute P(11) using the expression:

(9/5)(121) - (18/5)(11) + 14/5

Compute each term:

(9/5)(121) = 1089/5

(18/5)(11) = 198/5

So:

1089/5 - 198/5 + 14/5 = (1089 - 198 + 14)/5 = (1089 - 198 is 891, 891 +14 = 905)/5 = 905/5 = 181. Correct.

Therefore, the expression for P(x) is correct, so P(16) = 406.

But let me think again if there's another way to approach this problem, just to confirm.

Alternatively, since P(x) is a quadratic, and we have inequalities for all x, maybe we can consider that the quadratic P(x) must touch both bounds at some points. Since the difference between U(x) and L(x) is (x - 1)^2, which is zero only at x = 1. Therefore, since P(x) is squeezed between them, and at x = 1, both bounds are equal, so P(1) must be equal to that value. Let me check that.

Earlier, we saw that P(1) = 1. Let me check with the expression: (9/5)(1) - (18/5)(1) + 14/5 = (9 - 18 + 14)/5 = 5/5 = 1. Correct.

But maybe there's another point where P(x) touches one of the bounds? For example, if P(x) is tangent to L(x) or U(x) at some point. However, since the difference between P(x) and L(x) is a quadratic which is non-negative, and since the difference between U(x) and P(x) is also non-negative, then if P(x) is not equal to L(x) or U(x) everywhere, then the differences must be positive definite. However, since we found that P(x) is a convex combination of L(x) and U(x), which are both quadratics, and since the difference between them is a perfect square, then P(x) is determined uniquely by the value at one point? Wait, but we had to use P(11) = 181 to find k.

Alternatively, maybe there's another way to think about this. Let me consider that since P(x) is between L(x) and U(x) for all x, then the minimum of P(x) must be at least the minimum of L(x), and the maximum of P(x) must be at most the maximum of U(x). But since both L(x) and U(x) are quadratics, their minima and maxima can be found.

But L(x) = x² - 2x + 2. The minimum occurs at x = -b/(2a) = 2/(2*1) = 1. So minimum value is L(1) = 1 - 2 + 2 = 1. Similarly, U(x) = 2x² - 4x + 3. The minimum occurs at x = 4/(2*2) = 1. The minimum value is U(1) = 2 - 4 + 3 = 1. So both L(x) and U(x) have their minimum at x = 1, and both have the same minimum value of 1. Therefore, since P(x) is between them for all x, then P(x) must also have its minimum at x = 1, and P(1) = 1.

Wait, but if P(x) is between L(x) and U(x) for all x, then since L(x) and U(x) both have their minimum at x = 1, then P(x) must also have its minimum at x = 1? Because if P(x) were to have a minimum lower than 1, then it would violate the lower bound. Similarly, if P(x) had a maximum higher than the maximum of U(x), but since U(x) tends to infinity as x approaches infinity, and P(x) is a quadratic, if the leading coefficient of P(x) is 2, then it would have the same behavior as U(x). But since the leading coefficient of P(x) is between 1 and 2, as we found earlier, then P(x) would have a minimum at x = -b/(2a). Wait, but if P(x) is between L(x) and U(x), which both have their minima at x = 1, then maybe P(x) also has its minimum at x = 1?

Let me check. Suppose that P(x) has its minimum at x = 1. Then, since P(1) = 1, as we saw earlier. Let me check if that's necessarily true.

Suppose that P(x) is between L(x) and U(x) for all x. Since L(x) and U(x) both have their minimum at x = 1, and P(x) is between them, then P(x) must also have its minimum at x = 1. Because if P(x) had a minimum at some other point, then near that point, P(x) would be lower than L(x) or higher than U(x)? Wait, not necessarily. Wait, suppose that P(x) has a minimum at some other point. For example, suppose that P(x) has a minimum at x = c ≠ 1. Then, since L(x) and U(x) have their minima at x = 1, then depending on the position of c, maybe P(x) could be squeezed between them. But since both L(x) and U(x) have their minima at x = 1, and P(x) is between them, then if P(x) has a minimum at x = c, then at x = 1, P(1) must be at least L(1) = 1. But since P(x) is between L(x) and U(x), then P(1) must be equal to 1? Because L(1) = U(1) = 1, so P(1) must be equal to 1. Therefore, regardless of where the minimum of P(x) is, P(1) = 1.

But if P(x) has its minimum at x = 1, then that would make sense. However, if P(x) has its minimum at another point, then P(1) would still be 1. But how can that be? Let me think. Suppose that P(x) has its minimum at x = c ≠ 1. Then, since P(x) ≥ L(x) for all x, then at x = 1, P(1) ≥ L(1) = 1. But also, since P(x) ≤ U(x) for all x, then P(1) ≤ U(1) = 1. Therefore, P(1) must equal 1. Therefore, regardless of where the minimum of P(x) is, P(1) = 1.

Therefore, P(1) = 1. So, if I can find the coefficients of P(x) such that P(1) = 1, and also P(11) = 181, and the leading coefficient is between 1 and 2. But how does that help?

Alternatively, since we already found P(x) by assuming it's a convex combination of L(x) and U(x), and that worked, and gave us P(16) = 406, which seems correct.

But let me try another approach. Let me suppose that P(x) is a quadratic polynomial such that P(x) - L(x) is a non-negative quadratic and U(x) - P(x) is also a non-negative quadratic. Therefore, as we said before, P(x) - L(x) = (a - 1)x² + (b + 2)x + (c - 2) must be non-negative for all x. Similarly, U(x) - P(x) = (2 - a)x² + (-4 - b)x + (3 - c) must be non-negative for all x.

Since both quadratics are non-negative for all x, their leading coefficients are non-negative, so 1 ≤ a ≤ 2. Also, their discriminants must be ≤ 0.

So, for P(x) - L(x):

Discriminant D1 = (b + 2)^2 - 4*(a - 1)*(c - 2) ≤ 0

For U(x) - P(x):

Discriminant D2 = (-4 - b)^2 - 4*(2 - a)*(3 - c) ≤ 0

But we also know that P(x) is a quadratic, so we have three coefficients a, b, c. However, we have only two conditions from the discriminants, and also the value P(11) = 181. So, maybe we need more equations.

But we also know that P(1) = 1, as we saw earlier. Because P(1) is between L(1) and U(1), which are both 1, so P(1) must be 1. Therefore, P(1) = a*(1)^2 + b*(1) + c = a + b + c = 1.

So, we have three equations:

1. a + b + c = 1 (from P(1) = 1)
2. P(11) = 181: a*(121) + b*(11) + c = 181
3. The discriminant conditions D1 ≤ 0 and D2 ≤ 0.

But since we have two discriminant conditions, which are inequalities, but maybe we can find a, b, c by assuming equality in the discriminants? Because if the difference between P(x) and L(x) is a perfect square, then discriminant D1 = 0, and similarly for D2. But since we already found a solution by assuming P(x) is a convex combination of L(x) and U(x), which gives us a specific a, b, c, maybe that is the only solution?

Alternatively, since we have two inequalities (D1 ≤ 0 and D2 ≤ 0) and two equations (from P(1) = 1 and P(11) = 181), maybe we can solve for a, b, c.

But this seems complicated. However, since we already have a solution, maybe that is the unique solution? Let me check.

From our previous approach, we found that P(x) = (9/5)x² - (18/5)x + 14/5. Let me check the discriminant conditions.

First, compute P(x) - L(x):

P(x) - L(x) = [(9/5)x² - (18/5)x + 14/5] - [x² - 2x + 2] = (9/5 - 1)x² + (-18/5 + 2)x + (14/5 - 2) = (4/5)x² + (-18/5 + 10/5)x + (14/5 - 10/5) = (4/5)x² - (8/5)x + 4/5 = (4/5)(x² - 2x + 1) = (4/5)(x - 1)^2.

So discriminant D1 = [(-8/5)^2 - 4*(4/5)*(4/5)]? Wait, no. Wait, the discriminant of P(x) - L(x) is for the quadratic (4/5)x² - (8/5)x + 4/5.

So discriminant D1 = [(-8/5)^2 - 4*(4/5)*(4/5)] = (64/25) - (64/25) = 0. So discriminant is zero, which satisfies D1 ≤ 0. Similarly, compute U(x) - P(x):

U(x) - P(x) = [2x² - 4x + 3] - [(9/5)x² - (18/5)x + 14/5] = (2 - 9/5)x² + (-4 + 18/5)x + (3 - 14/5) = (1/5)x² + (-2/5)x + (1/5) = (1/5)(x² - 2x + 1) = (1/5)(x - 1)^2.

So discriminant D2 = [(-2/5)^2 - 4*(1/5)*(1/5)] = (4/25) - (4/25) = 0. So discriminant is zero, which satisfies D2 ≤ 0.

Therefore, our solution satisfies both discriminant conditions with equality. Therefore, this is the unique quadratic polynomial that touches both bounds at x = 1 (since discriminant zero implies that the quadratic touches the x-axis at one point, which is x = 1 here). Therefore, this must be the unique solution. Therefore, P(16) = 406.

Therefore, I think the answer is 406.

But just to make sure, let me check if there could be another quadratic that satisfies the inequalities and P(11) = 181. Suppose there is another quadratic. However, since the difference between U(x) and L(x) is (x - 1)^2, and we have P(x) lying between them, then P(x) must be equal to L(x) + k*(x - 1)^2 for some k between 0 and 1. Since we have P(11) = 181, which gives us k = 4/5, which is unique. Therefore, there is only one such quadratic. Therefore, P(16) is uniquely determined as 406.

Therefore, I think the answer is 406.

**Final Answer**
\boxed{406}

To solve the problem, we are given that $ P(x) $ is a quadratic polynomial satisfying the inequality:

$$
x^2 - 2x + 2 \leq P(x) \leq 2x^2 - 4x + 3 \quad \text{for all real } x,
$$
and that $ P(11) = 181 $. We are to find $ P(16) $.

---

### Step 1: Understand the Bounds

Let us define:
- $ L(x) = x^2 - 2x + 2 $
- $ U(x) = 2x^2 - 4x + 3 $

We are told that $ L(x) \leq P(x) \leq U(x) $ for all real $ x $, and that $ P(x) $ is a quadratic polynomial. This implies that $ P(x) $ lies between two quadratics, and since both bounds are quadratics, $ P(x) $ must also be a quadratic.

---

### Step 2: Analyze the Difference Between Bounds

Compute the difference between the upper and lower bounds:

$$
U(x) - L(x) = (2x^2 - 4x + 3) - (x^2 - 2x + 2) = x^2 - 2x + 1 = (x - 1)^2
$$

This is a perfect square, and it is always non-negative. Therefore, the difference between the upper and lower bounds is a perfect square, and the inequality $ L(x) \leq P(x) \leq U(x) $ suggests that $ P(x) $ can be written as a convex combination of $ L(x) $ and $ U(x) $, i.e.,

$$
P(x) = L(x) + k(x - 1)^2, \quad \text{where } 0 \leq k \leq 1
$$

This form ensures that $ P(x) $ lies between $ L(x) $ and $ U(x) $ for all $ x $, and the value of $ k $ determines how "close" $ P(x) $ is to the lower or upper bound.

---

### Step 3: Use the Given Value to Solve for $ k $

We are given that $ P(11) = 181 $. Let's compute $ P(11) $ using the expression:

$$
P(x) = L(x) + k(x - 1)^2
$$

First, compute $ L(11) $:

$$
L(11) = 11^2 - 2 \cdot 11 + 2 = 121 - 22 + 2 = 101
$$

Then compute $ (11 - 1)^2 = 100 $, so:

$$
P(11) = 101 + 100k = 181
$$

Solving for $ k $:

$$
100k = 80 \Rightarrow k = \frac{4}{5}
$$

---

### Step 4: Write the Full Expression for $ P(x) $

Now that we know $ k = \frac{4}{5} $, we can write:

$$
P(x) = L(x) + \frac{4}{5}(x - 1)^2
$$

Substitute $ L(x) = x^2 - 2x + 2 $ and expand:

$$
P(x) = x^2 - 2x + 2 + \frac{4}{5}(x^2 - 2x + 1)
$$

$$
= x^2 - 2x + 2 + \frac{4}{5}x^2 - \frac{8}{5}x + \frac{4}{5}
$$

Combine like terms:

- $ x^2 + \frac{4}{5}x^2 = \frac{9}{5}x^2 $
- $ -2x - \frac{8}{5}x = -\frac{18}{5}x $
- $ 2 + \frac{4}{5} = \frac{14}{5} $

So:

$$
P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}
$$

---

### Step 5: Compute $ P(16) $

Now compute $ P(16) $:

$$
P(16) = \frac{9}{5}(16)^2 - \frac{18}{5}(16) + \frac{14}{5}
$$

$$
= \frac{9}{5}(256) - \frac{18}{5}(16) + \frac{14}{5}
$$

$$
= \frac{2304}{5} - \frac{288}{5} + \frac{14}{5}
$$

$$
= \frac{2304 - 288 + 14}{5} = \frac{2030}{5} = 406
$$

---

### Final Answer

$$
\boxed{406}
$$

[Assistant]
Explanation:
Okay, so I need to find P(16) where P(x) is a quadratic polynomial that's squeezed between two other quadratics: x² - 2x + 2 and 2x² - 4x + 3 for all real numbers x. Also, they told me that P(11) = 181. Hmm, let me think about how to approach this.

First, since P(x) is a quadratic polynomial, I can write it in the general form: P(x) = ax² + bx + c, where a, b, c are real numbers. The problem says that for all real x, x² - 2x + 2 ≤ P(x) ≤ 2x² - 4x + 3. So, this inequality must hold for every x.

Since both the lower and upper bounds are quadratics, maybe P(x) is a quadratic that lies between them. But how can I find the coefficients a, b, c? Well, maybe I can consider the difference between P(x) and the lower bound, and the difference between the upper bound and P(x). Let me try that.

Let me denote the lower bound as L(x) = x² - 2x + 2 and the upper bound as U(x) = 2x² - 4x + 3. Then, for all x, L(x) ≤ P(x) ≤ U(x).

Since P(x) is a quadratic, maybe the difference between P(x) and L(x) is always non-negative, and the difference between U(x) and P(x) is also always non-negative. Let me write those differences:

P(x) - L(x) = (ax² + bx + c) - (x² - 2x + 2) = (a - 1)x² + (b + 2)x + (c - 2)

Similarly, U(x) - P(x) = (2x² - 4x + 3) - (ax² + bx + c) = (2 - a)x² + (-4 - b)x + (3 - c)

Since these differences must be non-negative for all real x, that means that each of these quadratics must be non-negative for all x. For a quadratic to be non-negative for all real x, it must satisfy two conditions: the leading coefficient is non-negative, and the discriminant is less than or equal to zero.

So, for P(x) - L(x) to be non-negative for all x:

1. The leading coefficient (a - 1) must be ≥ 0 ⇒ a ≥ 1
2. The discriminant must be ≤ 0. Let me compute the discriminant for P(x) - L(x):

Discriminant D1 = (b + 2)² - 4*(a - 1)*(c - 2) ≤ 0

Similarly, for U(x) - P(x) to be non-negative for all x:

1. The leading coefficient (2 - a) must be ≥ 0 ⇒ 2 - a ≥ 0 ⇒ a ≤ 2
2. The discriminant must be ≤ 0. Let me compute the discriminant for U(x) - P(x):

Discriminant D2 = (-4 - b)² - 4*(2 - a)*(3 - c) ≤ 0

So, combining the two inequalities from the leading coefficients, we have 1 ≤ a ≤ 2. Therefore, a is between 1 and 2.

Now, since P(x) is squeezed between L(x) and U(x) for all x, maybe P(x) is a linear combination of L(x) and U(x)? Or perhaps P(x) touches both bounds at some points? Wait, but since it's a quadratic, maybe it's tangent to both bounds? Or maybe it's equal to both bounds at some points?

Alternatively, since the inequalities hold for all x, maybe the difference between P(x) and L(x) is a non-negative quadratic, and the difference between U(x) and P(x) is also a non-negative quadratic. Therefore, both quadratics must be perfect squares? Or maybe they have a double root?

But maybe there's another approach. Let me think. If P(x) is between L(x) and U(x) for all x, then perhaps P(x) is equal to L(x) + something and U(x) - something else. But since both L(x) and U(x) are quadratics, maybe P(x) is a convex combination of them? Wait, but convex combinations would require coefficients that sum to 1, but here we don't know.

Alternatively, since both inequalities must hold for all x, maybe the minimum of U(x) - P(x) is zero and the maximum of P(x) - L(x) is zero? Wait, but that might not necessarily be the case. However, if P(x) is squeezed between L(x) and U(x) for all x, then maybe P(x) is equal to L(x) at some points and equal to U(x) at some other points? But since both L(x) and U(x) are quadratics, and P(x) is a quadratic, maybe they intersect at two points?

Alternatively, since the inequalities hold for all x, then the difference between P(x) and L(x) is non-negative for all x, and the difference between U(x) and P(x) is non-negative for all x. Therefore, both quadratics P(x) - L(x) and U(x) - P(x) are non-negative definite. Therefore, they must be perfect squares? Or at least have discriminant zero?

Wait, but if a quadratic is non-negative for all x, it can be written as k*(quadratic with leading coefficient positive) where the discriminant is zero? Or maybe not necessarily perfect squares. For example, if the discriminant is negative, then it's always positive. But if discriminant is zero, then it's a perfect square. However, since we have two different quadratics, maybe they have different discriminants?

But maybe there's a way to find P(x) by considering that the inequalities hold for all x. Let me think. Let me suppose that P(x) is equal to L(x) + k*(U(x) - L(x)) for some constant k between 0 and 1. Wait, that might be a way to express P(x) as a convex combination of L(x) and U(x). Let me check.

Let me compute U(x) - L(x):

U(x) - L(x) = (2x² - 4x + 3) - (x² - 2x + 2) = x² - 2x + 1 = (x - 1)^2.

Oh! That's interesting. So the difference between the upper and lower bounds is a perfect square. Therefore, U(x) - L(x) = (x - 1)^2. Therefore, if I write P(x) as L(x) + k*(x - 1)^2, where k is between 0 and 1, then P(x) would lie between L(x) and U(x). Because if k is 0, P(x) = L(x); if k is 1, P(x) = U(x). And for 0 < k < 1, P(x) is between them.

But since P(x) is a quadratic, and L(x) is quadratic, adding k*(x - 1)^2 would still give a quadratic. Therefore, maybe P(x) is of the form L(x) + k*(x - 1)^2. Let me check if this works.

Let me write that down:

P(x) = L(x) + k*(x - 1)^2 = x² - 2x + 2 + k*(x² - 2x + 1) = (1 + k)x² + (-2 - 2k)x + (2 + k)

Alternatively, since U(x) = L(x) + (x - 1)^2, so P(x) = L(x) + k*(x - 1)^2, where 0 ≤ k ≤ 1.

But then, since P(x) is a quadratic, and we have P(11) = 181, we can use that to find k. Let me try that.

So, let me compute P(11):

P(11) = (1 + k)*(11)^2 + (-2 - 2k)*(11) + (2 + k)

Let me compute each term:

First term: (1 + k)*121 = 121 + 121k

Second term: (-2 - 2k)*11 = -22 - 22k

Third term: 2 + k

Adding them all together:

121 + 121k - 22 - 22k + 2 + k = (121 - 22 + 2) + (121k - 22k + k) = 101 + 100k

So, P(11) = 101 + 100k. But we are told that P(11) = 181. Therefore:

101 + 100k = 181 ⇒ 100k = 80 ⇒ k = 80 / 100 = 0.8 = 4/5.

So, k = 4/5. Therefore, P(x) = L(x) + (4/5)*(x - 1)^2. Let me write that out:

P(x) = x² - 2x + 2 + (4/5)(x² - 2x + 1) = Let me compute that:

First, expand (4/5)(x² - 2x + 1) = (4/5)x² - (8/5)x + 4/5

Then add to L(x):

x² - 2x + 2 + (4/5)x² - (8/5)x + 4/5

Combine like terms:

x² + (4/5)x² = (9/5)x²

-2x - (8/5)x = (-10/5 - 8/5)x = (-18/5)x

2 + 4/5 = 10/5 + 4/5 = 14/5

Therefore, P(x) = (9/5)x² - (18/5)x + 14/5.

Alternatively, we can factor out 1/5:

P(x) = (1/5)(9x² - 18x + 14).

But let me check if this is correct. Let me verify if this P(x) satisfies the original inequalities.

First, check if P(x) is between L(x) and U(x) for all x. Since we constructed it as L(x) + k*(x - 1)^2 with 0 < k < 1, then yes, it should be between them. But let me check for a specific value. For example, at x = 1, L(1) = 1 - 2 + 2 = 1, U(1) = 2 - 4 + 3 = 1. So both bounds are equal to 1 at x = 1. Then P(1) = (9/5)(1) - (18/5)(1) + 14/5 = (9 - 18 + 14)/5 = 5/5 = 1. So P(1) = 1, which is equal to both bounds. That makes sense because the difference between U and L is (x - 1)^2, so at x = 1, the difference is zero, so P(x) must be equal to both bounds there.

Another check: Let me check x = 0. L(0) = 0 - 0 + 2 = 2. U(0) = 0 - 0 + 3 = 3. P(0) = 0 - 0 + 14/5 = 14/5 = 2.8, which is between 2 and 3. Good.

Another check: Let me check x = 2. L(2) = 4 - 4 + 2 = 2. U(2) = 8 - 8 + 3 = 3. P(2) = (9/5)(4) - (18/5)(2) + 14/5 = 36/5 - 36/5 + 14/5 = 14/5 = 2.8. Again between 2 and 3.

So seems okay. Now, since we found k = 4/5, then P(x) is determined. Therefore, we can compute P(16) using this expression.

So, let me compute P(16):

P(16) = (9/5)(16)^2 - (18/5)(16) + 14/5

First compute each term:

(9/5)(256) = (9 * 256)/5

(18/5)(16) = (18 * 16)/5

So:

P(16) = (9*256 - 18*16 + 14)/5

Let me compute numerator:

First, compute 9*256:

Well, 256 * 9: 256*10 = 2560, minus 256 = 2560 - 256 = 2304

Then 18*16 = 288

So numerator is 2304 - 288 + 14 = Let's compute 2304 - 288 first:

2304 - 288: 2304 - 200 = 2104; 2104 - 88 = 2016

Then 2016 + 14 = 2030

Therefore, P(16) = 2030 / 5 = 406.

Wait, 2030 divided by 5 is 406? Let me check:

5 * 400 = 2000, so 2030 - 2000 = 30, so 30 / 5 = 6. Therefore, 400 + 6 = 406. Yes, correct.

So P(16) = 406. Therefore, the answer is 406.

But let me check again if my expression for P(x) is correct. Let me verify with another point. For example, let me check P(11) = 181.

Earlier, we had P(11) = 101 + 100k, and with k = 4/5, that gives 101 + 80 = 181. Correct. So that's correct.

Alternatively, compute P(11) using the expression:

(9/5)(121) - (18/5)(11) + 14/5

Compute each term:

(9/5)(121) = 1089/5

(18/5)(11) = 198/5

So:

1089/5 - 198/5 + 14/5 = (1089 - 198 + 14)/5 = (1089 - 198 is 891, 891 +14 = 905)/5 = 905/5 = 181. Correct.

Therefore, the expression for P(x) is correct, so P(16) = 406.

But let me think again if there's another way to approach this problem, just to confirm.

Alternatively, since P(x) is a quadratic, and we have inequalities for all x, maybe we can consider that the quadratic P(x) must touch both bounds at some points. Since the difference between U(x) and L(x) is (x - 1)^2, which is zero only at x = 1. Therefore, since P(x) is squeezed between them, and at x = 1, both bounds are equal, so P(1) must be equal to that value. Let me check that.

Earlier, we saw that P(1) = 1. Let me check with the expression: (9/5)(1) - (18/5)(1) + 14/5 = (9 - 18 + 14)/5 = 5/5 = 1. Correct.

But maybe there's another point where P(x) touches one of the bounds? For example, if P(x) is tangent to L(x) or U(x) at some point. However, since the difference between P(x) and L(x) is a quadratic which is non-negative, and since the difference between U(x) and P(x) is also non-negative, then if P(x) is not equal to L(x) or U(x) everywhere, then the differences must be positive definite. However, since we found that P(x) is a convex combination of L(x) and U(x), which are both quadratics, and since the difference between them is a perfect square, then P(x) is determined uniquely by the value at one point? Wait, but we had to use P(11) = 181 to find k.

Alternatively, maybe there's another way to think about this. Let me consider that since P(x) is between L(x) and U(x) for all x, then the minimum of P(x) must be at least the minimum of L(x), and the maximum of P(x) must be at most the maximum of U(x). But since both L(x) and U(x) are quadratics, their minima and maxima can be found.

But L(x) = x² - 2x + 2. The minimum occurs at x = -b/(2a) = 2/(2*1) = 1. So minimum value is L(1) = 1 - 2 + 2 = 1. Similarly, U(x) = 2x² - 4x + 3. The minimum occurs at x = 4/(2*2) = 1. The minimum value is U(1) = 2 - 4 + 3 = 1. So both L(x) and U(x) have their minimum at x = 1, and both have the same minimum value of 1. Therefore, since P(x) is between them for all x, then P(x) must also have its minimum at x = 1, and P(1) = 1.

Wait, but if P(x) is between L(x) and U(x) for all x, then since L(x) and U(x) both have their minimum at x = 1, then P(x) must also have its minimum at x = 1? Because if P(x) were to have a minimum lower than 1, then it would violate the lower bound. Similarly, if P(x) had a maximum higher than the maximum of U(x), but since U(x) tends to infinity as x approaches infinity, and P(x) is a quadratic, if the leading coefficient of P(x) is 2, then it would have the same behavior as U(x). But since the leading coefficient of P(x) is between 1 and 2, as we found earlier, then P(x) would have a minimum at x = -b/(2a). Wait, but if P(x) is between L(x) and U(x), which both have their minima at x = 1, then maybe P(x) also has its minimum at x = 1?

Let me check. Suppose that P(x) has its minimum at x = 1. Then, since P(1) = 1, as we saw earlier. Let me check if that's necessarily true.

Suppose that P(x) is between L(x) and U(x) for all x. Since L(x) and U(x) both have their minimum at x = 1, and P(x) is between them, then P(x) must also have its minimum at x = 1. Because if P(x) had a minimum at some other point, then near that point, P(x) would be lower than L(x) or higher than U(x)? Wait, not necessarily. Wait, suppose that P(x) has a minimum at some other point. For example, suppose that P(x) has a minimum at x = c ≠ 1. Then, since L(x) and U(x) have their minima at x = 1, then depending on the position of c, maybe P(x) could be squeezed between them. But since both L(x) and U(x) have their minima at x = 1, and P(x) is between them, then if P(x) has a minimum at x = c, then at x = 1, P(1) must be at least L(1) = 1. But since P(x) is between L(x) and U(x), then P(1) must be equal to 1? Because L(1) = U(1) = 1, so P(1) must be equal to 1. Therefore, regardless of where the minimum of P(x) is, P(1) = 1.

But if P(x) has its minimum at x = 1, then that would make sense. However, if P(x) has its minimum at another point, then P(1) would still be 1. But how can that be? Let me think. Suppose that P(x) has its minimum at x = c ≠ 1. Then, since P(x) ≥ L(x) for all x, then at x = 1, P(1) ≥ L(1) = 1. But also, since P(x) ≤ U(x) for all x, then P(1) ≤ U(1) = 1. Therefore, P(1) must equal 1. Therefore, regardless of where the minimum of P(x) is, P(1) = 1.

Therefore, P(1) = 1. So, if I can find the coefficients of P(x) such that P(1) = 1, and also P(11) = 181, and the leading coefficient is between 1 and 2. But how does that help?

Alternatively, since we already found P(x) by assuming it's a convex combination of L(x) and U(x), and that worked, and gave us P(16) = 406, which seems correct.

But let me try another approach. Let me suppose that P(x) is a quadratic polynomial such that P(x) - L(x) is a non-negative quadratic and U(x) - P(x) is also a non-negative quadratic. Therefore, as we said before, P(x) - L(x) = (a - 1)x² + (b + 2)x + (c - 2) must be non-negative for all x. Similarly, U(x) - P(x) = (2 - a)x² + (-4 - b)x + (3 - c) must be non-negative for all x.

Since both quadratics are non-negative for all x, their leading coefficients are non-negative, so 1 ≤ a ≤ 2. Also, their discriminants must be ≤ 0.

So, for P(x) - L(x):

Discriminant D1 = (b + 2)^2 - 4*(a - 1)*(c - 2) ≤ 0

For U(x) - P(x):

Discriminant D2 = (-4 - b)^2 - 4*(2 - a)*(3 - c) ≤ 0

But we also know that P(x) is a quadratic, so we have three coefficients a, b, c. However, we have only two conditions from the discriminants, and also the value P(11) = 181. So, maybe we need more equations.

But we also know that P(1) = 1, as we saw earlier. Because P(1) is between L(1) and U(1), which are both 1, so P(1) must be 1. Therefore, P(1) = a*(1)^2 + b*(1) + c = a + b + c = 1.

So, we have three equations:

1. a + b + c = 1 (from P(1) = 1)
2. P(11) = 181: a*(121) + b*(11) + c = 181
3. The discriminant conditions D1 ≤ 0 and D2 ≤ 0.

But since we have two discriminant conditions, which are inequalities, but maybe we can find a, b, c by assuming equality in the discriminants? Because if the difference between P(x) and L(x) is a perfect square, then discriminant D1 = 0, and similarly for D2. But since we already found a solution by assuming P(x) is a convex combination of L(x) and U(x), which gives us a specific a, b, c, maybe that is the only solution?

Alternatively, since we have two inequalities (D1 ≤ 0 and D2 ≤ 0) and two equations (from P(1) = 1 and P(11) = 181), maybe we can solve for a, b, c.

But this seems complicated. However, since we already have a solution, maybe that is the unique solution? Let me check.

From our previous approach, we found that P(x) = (9/5)x² - (18/5)x + 14/5. Let me check the discriminant conditions.

First, compute P(x) - L(x):

P(x) - L(x) = [(9/5)x² - (18/5)x + 14/5] - [x² - 2x + 2] = (9/5 - 1)x² + (-18/5 + 2)x + (14/5 - 2) = (4/5)x² + (-18/5 + 10/5)x + (14/5 - 10/5) = (4/5)x² - (8/5)x + 4/5 = (4/5)(x² - 2x + 1) = (4/5)(x - 1)^2.

So discriminant D1 = [(-8/5)^2 - 4*(4/5)*(4/5)]? Wait, no. Wait, the discriminant of P(x) - L(x) is for the quadratic (4/5)x² - (8/5)x + 4/5.

So discriminant D1 = [(-8/5)^2 - 4*(4/5)*(4/5)] = (64/25) - (64/25) = 0. So discriminant is zero, which satisfies D1 ≤ 0. Similarly, compute U(x) - P(x):

U(x) - P(x) = [2x² - 4x + 3] - [(9/5)x² - (18/5)x + 14/5] = (2 - 9/5)x² + (-4 + 18/5)x + (3 - 14/5) = (1/5)x² + (-2/5)x + (1/5) = (1/5)(x² - 2x + 1) = (1/5)(x - 1)^2.

So discriminant D2 = [(-2/5)^2 - 4*(1/5)*(1/5)] = (4/25) - (4/25) = 0. So discriminant is zero, which satisfies D2 ≤ 0.

Therefore, our solution satisfies both discriminant conditions with equality. Therefore, this is the unique quadratic polynomial that touches both bounds at x = 1 (since discriminant zero implies that the quadratic touches the x-axis at one point, which is x = 1 here). Therefore, this must be the unique solution. Therefore, P(16) = 406.

Therefore, I think the answer is 406.

But just to make sure, let me check if there could be another quadratic that satisfies the inequalities and P(11) = 181. Suppose there is another quadratic. However, since the difference between U(x) and L(x) is (x - 1)^2, and we have P(x) lying between them, then P(x) must be equal to L(x) + k*(x - 1)^2 for some k between 0 and 1. Since we have P(11) = 181, which gives us k = 4/5, which is unique. Therefore, there is only one such quadratic. Therefore, P(16) is uniquely determined as 406.

Therefore, I think the answer is 406.

**Final Answer**
\boxed{406}

To solve the problem, we are given that $ P(x) $ is a quadratic polynomial satisfying the inequality:

$$
x^2 - 2x + 2 \leq P(x) \leq 2x^2 - 4x + 3 \quad \text{for all real } x,
$$
and that $ P(11) = 181 $. We are to find $ P(16) $.

---

### Step 1: Understand the Bounds

Let us define:
- $ L(x) = x^2 - 2x + 2 $
- $ U(x) = 2x^2 - 4x + 3 $

We are told that $ L(x) \leq P(x) \leq U(x) $ for all real $ x $, and that $ P(x) $ is a quadratic polynomial. This implies that $ P(x) $ lies between two quadratics, and since both bounds are quadratics, $ P(x) $ must also be a quadratic.

---

### Step 2: Analyze the Difference Between Bounds

Compute the difference between the upper and lower bounds:

$$
U(x) - L(x) = (2x^2 - 4x + 3) - (x^2 - 2x + 2) = x^2 - 2x + 1 = (x - 1)^2
$$

This is a perfect square, and it is always non-negative. Therefore, the difference between the upper and lower bounds is a perfect square, and the inequality $ L(x) \leq P(x) \leq U(x) $ suggests that $ P(x) $ can be written as a convex combination of $ L(x) $ and $ U(x) $, i.e.,

$$
P(x) = L(x) + k(x - 1)^2, \quad \text{where } 0 \leq k \leq 1
$$

This form ensures that $ P(x) $ lies between $ L(x) $ and $ U(x) $ for all $ x $, and the value of $ k $ determines how "close" $ P(x) $ is to the lower or upper bound.

---

### Step 3: Use the Given Value to Solve for $ k $

We are given that $ P(11) = 181 $. Let's compute $ P(11) $ using the expression:

$$
P(x) = L(x) + k(x - 1)^2
$$

First, compute $ L(11) $:

$$
L(11) = 11^2 - 2 \cdot 11 + 2 = 121 - 22 + 2 = 101
$$

Then compute $ (11 - 1)^2 = 100 $, so:

$$
P(11) = 101 + 100k = 181
$$

Solving for $ k $:

$$
100k = 80 \Rightarrow k = \frac{4}{5}
$$

---

### Step 4: Write the Full Expression for $ P(x) $

Now that we know $ k = \frac{4}{5} $, we can write:

$$
P(x) = L(x) + \frac{4}{5}(x - 1)^2
$$

Substitute $ L(x) = x^2 - 2x + 2 $ and expand:

$$
P(x) = x^2 - 2x + 2 + \frac{4}{5}(x^2 - 2x + 1)
$$

$$
= x^2 - 2x + 2 + \frac{4}{5}x^2 - \frac{8}{5}x + \frac{4}{5}
$$

Combine like terms:

- $ x^2 + \frac{4}{5}x^2 = \frac{9}{5}x^2 $
- $ -2x - \frac{8}{5}x = -\frac{18}{5}x $
- $ 2 + \frac{4}{5} = \frac{14}{5} $

So:

$$
P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}
$$

---

### Step 5: Compute $ P(16) $

Now compute $ P(16) $:

$$
P(16) = \frac{9}{5}(16)^2 - \frac{18}{5}(16) + \frac{14}{5}
$$

$$
= \frac{9}{5}(256) - \frac{18}{5}(16) + \frac{14}{5}
$$

$$
= \frac{2304}{5} - \frac{288}{5} + \frac{14}{5}
$$

$$
= \frac{2304 - 288 + 14}{5} = \frac{2030}{5} = 406
$$

---

### Final Answer

$$
\boxed{406}
$$

Solution: Rewriting the given quadratics in vertex form, we have \[1 + (x-1)^2 \le P(x) \le 1 + 2(x-1)^2.\]Both of those quadratics have vertex at $(1, 1)$; considering the shape of the graph of a quadratic, we see that $P$ must also have its vertex at $(1,1)$. Therefore, \[P(x) = 1 + k(x-1)^2\]for some constant $k$. Setting $x = 11$, we have $181 = 1 +100k$, so $k = \tfrac{9}{5}$. Then \[P(16) = 1 + \tfrac{9}{5} \cdot 15^2 = \boxed{406}.\]


Explanation:
Okay, so I need to solve this equation: (3)5² - 4(5 - a)² ÷ 3 = 63. Hmm, let me make sure I understand the equation correctly. Let me rewrite it to be sure. It says 3 times 5 squared minus 4 times (5 minus a) squared divided by 3 equals 63. So, written out, it's 3*5² - [4*(5 - a)²]/3 = 63. Alright, that seems right.

First, maybe I should simplify the left side step by step. Let me start by calculating 3*5². Wait, 5 squared is 25, so 3*25 is 75. So the first term is 75. Then, the second term is 4*(5 - a)² divided by 3. So the entire equation becomes 75 - [4*(5 - a)²]/3 = 63.

Now, I need to solve for a. Let me rearrange the equation to isolate the term with a. Let me subtract 75 from both sides. That would give me - [4*(5 - a)²]/3 = 63 - 75. Calculating 63 - 75, that's -12. So now the equation is - [4*(5 - a)²]/3 = -12.

Hmm, maybe I can multiply both sides by -1 to make it simpler. That would give me [4*(5 - a)²]/3 = 12. Then, to get rid of the denominator, I can multiply both sides by 3. So 4*(5 - a)² = 36.

Now, divide both sides by 4 to solve for (5 - a)². That would be (5 - a)² = 36 / 4, which is 9. So (5 - a)² = 9.

To solve for (5 - a), take the square root of both sides. Remembering that square roots can be positive or negative, so 5 - a = ±√9, which is ±3. Therefore, 5 - a = 3 or 5 - a = -3.

Let me solve each equation separately.

First equation: 5 - a = 3. Subtract 5 from both sides: -a = 3 - 5, which is -2. Multiply both sides by -1: a = 2.

Second equation: 5 - a = -3. Subtract 5 from both sides: -a = -3 - 5, which is -8. Multiply both sides by -1: a = 8.

So the solutions are a = 2 and a = 8. The question asks for the sum of the values of a. So adding them together: 2 + 8 = 10.

Wait, let me check my steps again to make sure I didn't make a mistake.

Original equation: 3*5² - [4*(5 - a)²]/3 = 63.

Calculating 3*25 is 75. Then, 75 - [something] = 63. So [something] must be 75 - 63 = 12. Wait, but in my earlier steps, I had - [something] = -12, so [something] = 12. That seems correct.

Then, [4*(5 - a)²]/3 = 12. Multiply both sides by 3: 4*(5 - a)² = 36. Divide by 4: (5 - a)² = 9. Square root: 5 - a = ±3. Then solving for a gives 2 and 8.

Adding them gives 10. Let me verify by plugging the values back into the original equation.

First, a = 2.

Original equation: 3*5² - [4*(5 - 2)²]/3.

Calculate 5 - 2 = 3. Then 3 squared is 9. Multiply by 4: 36. Divide by 3: 12. Then 3*25 is 75. So 75 - 12 = 63. Which matches the right side.

Now, a = 8.

Original equation: 3*5² - [4*(5 - 8)²]/3.

5 - 8 = -3. Squared is 9. Multiply by 4: 36. Divide by 3: 12. Then 75 - 12 = 63. Also correct.

So both solutions work. Therefore, the sum is 2 + 8 = 10.

Wait, but let me check if there's another way to approach this problem, just to be thorough. Maybe expanding the terms? Let me try that.

Original equation: 3*5² - [4*(5 - a)²]/3 = 63.

Alternatively, maybe I can write the entire equation without breaking it down. Let me see.

But I think my initial approach was correct. However, just to make sure, let me try expanding (5 - a)².

(5 - a)² = 25 - 10a + a².

So substituting back into the equation:

3*25 - [4*(25 - 10a + a²)] / 3 = 63.

Calculating 3*25 is 75. Then, expanding the numerator: 4*(25 - 10a + a²) = 100 - 40a + 4a².

So the equation becomes 75 - [100 - 40a + 4a²]/3 = 63.

To eliminate the denominator, multiply every term by 3:

3*75 - (100 - 40a + 4a²) = 3*63.

Calculating each term: 3*75 is 225, 3*63 is 189.

So the equation is 225 - 100 + 40a - 4a² = 189.

Simplify the left side: 225 - 100 is 125, so 125 + 40a - 4a² = 189.

Rearranging terms: -4a² + 40a + 125 = 189.

Subtract 189 from both sides: -4a² + 40a + 125 - 189 = 0.

Calculating 125 - 189 is -64. So equation becomes -4a² + 40a - 64 = 0.

Multiply both sides by -1 to make the quadratic coefficient positive: 4a² - 40a + 64 = 0.

Divide all terms by 4: a² - 10a + 16 = 0.

Now, solving this quadratic equation. Let's factor it. Looking for two numbers that multiply to 16 and add to -10.

Factors of 16: 1 & 16, 2 & 8, 4 & 4.

Looking for two numbers that add to -10. Let's see, -2 and -8 multiply to 16 and add to -10.

So, the equation factors as (a - 2)(a - 8) = 0. Therefore, solutions are a = 2 and a = 8.

Same answers as before! So the sum is 2 + 8 = 10.

Therefore, regardless of the method, the sum of the values of a is 10.

I think that's thorough. I considered two different methods: one by simplifying step by step and solving for a, and another by expanding and solving the quadratic. Both led to the same solutions. So I'm confident that the answer is 10.

**Final Answer**
The sum of the values of $ a $ is \boxed{10}.

To solve the equation:

$$
(3)5^2 - \frac{4(5 - a)^2}{3} = 63
$$

we proceed step by step.

---

### Step 1: Simplify the known terms

First, compute $ 3 \cdot 5^2 $:

$$
3 \cdot 25 = 75
$$

So the equation becomes:

$$
75 - \frac{4(5 - a)^2}{3} = 63
$$

---

### Step 2: Isolate the squared term

Subtract 75 from both sides:

$$
- \frac{4(5 - a)^2}{3} = 63 - 75 = -12
$$

Multiply both sides by -1:

$$
\frac{4(5 - a)^2}{3} = 12
$$

Multiply both sides by 3:

$$
4(5 - a)^2 = 36
$$

Divide both sides by 4:

$$
(5 - a)^2 = 9
$$

---

### Step 3: Solve the quadratic equation

Take the square root of both sides:

$$
5 - a = \pm 3
$$

This gives two equations:

1. $ 5 - a = 3 \Rightarrow a = 2 $
2. $ 5 - a = -3 \Rightarrow a = 8 $

---

### Step 4: Find the sum of the solutions

The two values of $ a $ are 2 and 8. Their sum is:

$$
2 + 8 = \boxed{10}
$$

---

### Final Answer

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: (3)5² - 4(5 - a)² ÷ 3 = 63. Hmm, let me make sure I understand the equation correctly. Let me rewrite it to be sure. It says 3 times 5 squared minus 4 times (5 minus a) squared divided by 3 equals 63. So, written out, it's 3*5² - [4*(5 - a)²]/3 = 63. Alright, that seems right.

First, maybe I should simplify the left side step by step. Let me start by calculating 3*5². Wait, 5 squared is 25, so 3*25 is 75. So the first term is 75. Then, the second term is 4*(5 - a)² divided by 3. So the entire equation becomes 75 - [4*(5 - a)²]/3 = 63.

Now, I need to solve for a. Let me rearrange the equation to isolate the term with a. Let me subtract 75 from both sides. That would give me - [4*(5 - a)²]/3 = 63 - 75. Calculating 63 - 75, that's -12. So now the equation is - [4*(5 - a)²]/3 = -12.

Hmm, maybe I can multiply both sides by -1 to make it simpler. That would give me [4*(5 - a)²]/3 = 12. Then, to get rid of the denominator, I can multiply both sides by 3. So 4*(5 - a)² = 36.

Now, divide both sides by 4 to solve for (5 - a)². That would be (5 - a)² = 36 / 4, which is 9. So (5 - a)² = 9.

To solve for (5 - a), take the square root of both sides. Remembering that square roots can be positive or negative, so 5 - a = ±√9, which is ±3. Therefore, 5 - a = 3 or 5 - a = -3.

Let me solve each equation separately.

First equation: 5 - a = 3. Subtract 5 from both sides: -a = 3 - 5, which is -2. Multiply both sides by -1: a = 2.

Second equation: 5 - a = -3. Subtract 5 from both sides: -a = -3 - 5, which is -8. Multiply both sides by -1: a = 8.

So the solutions are a = 2 and a = 8. The question asks for the sum of the values of a. So adding them together: 2 + 8 = 10.

Wait, let me check my steps again to make sure I didn't make a mistake.

Original equation: 3*5² - [4*(5 - a)²]/3 = 63.

Calculating 3*25 is 75. Then, 75 - [something] = 63. So [something] must be 75 - 63 = 12. Wait, but in my earlier steps, I had - [something] = -12, so [something] = 12. That seems correct.

Then, [4*(5 - a)²]/3 = 12. Multiply both sides by 3: 4*(5 - a)² = 36. Divide by 4: (5 - a)² = 9. Square root: 5 - a = ±3. Then solving for a gives 2 and 8.

Adding them gives 10. Let me verify by plugging the values back into the original equation.

First, a = 2.

Original equation: 3*5² - [4*(5 - 2)²]/3.

Calculate 5 - 2 = 3. Then 3 squared is 9. Multiply by 4: 36. Divide by 3: 12. Then 3*25 is 75. So 75 - 12 = 63. Which matches the right side.

Now, a = 8.

Original equation: 3*5² - [4*(5 - 8)²]/3.

5 - 8 = -3. Squared is 9. Multiply by 4: 36. Divide by 3: 12. Then 75 - 12 = 63. Also correct.

So both solutions work. Therefore, the sum is 2 + 8 = 10.

Wait, but let me check if there's another way to approach this problem, just to be thorough. Maybe expanding the terms? Let me try that.

Original equation: 3*5² - [4*(5 - a)²]/3 = 63.

Alternatively, maybe I can write the entire equation without breaking it down. Let me see.

But I think my initial approach was correct. However, just to make sure, let me try expanding (5 - a)².

(5 - a)² = 25 - 10a + a².

So substituting back into the equation:

3*25 - [4*(25 - 10a + a²)] / 3 = 63.

Calculating 3*25 is 75. Then, expanding the numerator: 4*(25 - 10a + a²) = 100 - 40a + 4a².

So the equation becomes 75 - [100 - 40a + 4a²]/3 = 63.

To eliminate the denominator, multiply every term by 3:

3*75 - (100 - 40a + 4a²) = 3*63.

Calculating each term: 3*75 is 225, 3*63 is 189.

So the equation is 225 - 100 + 40a - 4a² = 189.

Simplify the left side: 225 - 100 is 125, so 125 + 40a - 4a² = 189.

Rearranging terms: -4a² + 40a + 125 = 189.

Subtract 189 from both sides: -4a² + 40a + 125 - 189 = 0.

Calculating 125 - 189 is -64. So equation becomes -4a² + 40a - 64 = 0.

Multiply both sides by -1 to make the quadratic coefficient positive: 4a² - 40a + 64 = 0.

Divide all terms by 4: a² - 10a + 16 = 0.

Now, solving this quadratic equation. Let's factor it. Looking for two numbers that multiply to 16 and add to -10.

Factors of 16: 1 & 16, 2 & 8, 4 & 4.

Looking for two numbers that add to -10. Let's see, -2 and -8 multiply to 16 and add to -10.

So, the equation factors as (a - 2)(a - 8) = 0. Therefore, solutions are a = 2 and a = 8.

Same answers as before! So the sum is 2 + 8 = 10.

Therefore, regardless of the method, the sum of the values of a is 10.

I think that's thorough. I considered two different methods: one by simplifying step by step and solving for a, and another by expanding and solving the quadratic. Both led to the same solutions. So I'm confident that the answer is 10.

**Final Answer**
The sum of the values of $ a $ is \boxed{10}.

To solve the equation:

$$
(3)5^2 - \frac{4(5 - a)^2}{3} = 63
$$

we proceed step by step.

---

### Step 1: Simplify the known terms

First, compute $ 3 \cdot 5^2 $:

$$
3 \cdot 25 = 75
$$

So the equation becomes:

$$
75 - \frac{4(5 - a)^2}{3} = 63
$$

---

### Step 2: Isolate the squared term

Subtract 75 from both sides:

$$
- \frac{4(5 - a)^2}{3} = 63 - 75 = -12
$$

Multiply both sides by -1:

$$
\frac{4(5 - a)^2}{3} = 12
$$

Multiply both sides by 3:

$$
4(5 - a)^2 = 36
$$

Divide both sides by 4:

$$
(5 - a)^2 = 9
$$

---

### Step 3: Solve the quadratic equation

Take the square root of both sides:

$$
5 - a = \pm 3
$$

This gives two equations:

1. $ 5 - a = 3 \Rightarrow a = 2 $
2. $ 5 - a = -3 \Rightarrow a = 8 $

---

### Step 4: Find the sum of the solutions

The two values of $ a $ are 2 and 8. Their sum is:

$$
2 + 8 = \boxed{10}
$$

---

### Final Answer

$$
\boxed{10}
$$

Solution:
First we multiply all terms in the equation by 3 to avoid fractions and then solve for $a$. \begin{align*}
9\cdot5^2-4(5-a)^2&=3\cdot63\quad\Rightarrow\\
-4(5-a)^2&=9\cdot21-9\cdot25\quad\Rightarrow\\
&=9(-4)\quad\Rightarrow\\
(5-a)^2&=9
\end{align*} Therefore, \begin{align*}
5-a=3\quad\text{ OR }\quad 5-a=-3\quad\Rightarrow\\
2=a \quad\text{ OR }\quad 8=a.
\end{align*} The sum of the values of $a$ is $2+8=\boxed{10}$.


Explanation:
Okay, so I need to find a vector v such that when I take the cross product of the vector (1, 2, -5) with v, I get the vector (90, 30, 30). And among all possible such vectors v, I need to find the one with the smallest magnitude. Hmm, cross products... Let me recall how cross products work.

First, the cross product of two vectors in three dimensions results in a vector that's perpendicular to both of the original vectors. So, if I have vectors a and b, then a × b is perpendicular to both a and b. That means that the cross product here, which is (1, 2, -5) × v = (90, 30, 30), must be perpendicular to both (1, 2, -5) and v. Wait, but maybe that's not directly helpful here. Let me think.

Alternatively, maybe I can set up equations. Let me denote vector v as (x, y, z). Then, the cross product of (1, 2, -5) and (x, y, z) is given by the determinant of the following matrix:

|i   j   k |
|1    2  -5|
|x    y   z|

Calculating that determinant, the cross product should be:

i*(2*z - (-5)*y) - j*(1*z - (-5)*x) + k*(1*y - 2*x)

Simplifying each component:

First component (i): 2z + 5y

Second component (j): -(z + 5x) → -z -5x

Third component (k): y - 2x

So, the cross product is (2z + 5y, -z -5x, y - 2x). According to the problem, this equals (90, 30, 30). Therefore, we can set up the following system of equations:

1) 2z + 5y = 90

2) -z -5x = 30

3) y - 2x = 30

So, three equations with three variables x, y, z. Let me write them again:

Equation 1: 5y + 2z = 90

Equation 2: -5x - z = 30

Equation 3: -2x + y = 30

So, I need to solve this system. Let me try solving equations 2 and 3 first for x and y, and then substitute into equation 1 to find z.

From equation 3: y = 2x + 30. Let me write that down.

Equation 3: y = 2x + 30

Now, substitute this into equation 2: -5x - z = 30. Wait, equation 2 is -5x - z = 30. So, if I can express z in terms of x, maybe?

Let me rearrange equation 2: -z = 30 + 5x → z = -30 -5x.

So z = -5x -30.

Now, substitute y and z into equation 1: 5y + 2z = 90.

Substituting y = 2x + 30 and z = -5x -30:

5*(2x + 30) + 2*(-5x -30) = 90

Let me compute that:

First term: 5*(2x + 30) = 10x + 150

Second term: 2*(-5x -30) = -10x -60

Adding them together: 10x + 150 -10x -60 = (10x -10x) + (150 -60) = 0x + 90 = 90

So, 90 = 90. Wait, that's an identity. So, that means that equations 1, 2, 3 are dependent? So, there are infinitely many solutions? Which matches the problem statement that there are infinitely many vectors v satisfying the equation. Therefore, we need to find the vector v with the smallest magnitude.

So, since equations 2 and 3 give us expressions for y and z in terms of x, we can write v as (x, y, z) = (x, 2x + 30, -5x -30). Therefore, the vector v is parameterized by x. So, to find the vector with the smallest magnitude, we need to minimize the magnitude of v as a function of x.

So, let me write the magnitude squared of v, since minimizing the magnitude is equivalent to minimizing the magnitude squared, which is easier.

Let me denote ||v||² = x² + y² + z².

Substituting y and z:

||v||² = x² + (2x + 30)² + (-5x -30)²

Let me expand each term:

First term: x²

Second term: (2x + 30)² = 4x² + 120x + 900

Third term: (-5x -30)² = (5x + 30)² = 25x² + 300x + 900

Adding them all together:

x² + 4x² + 120x + 900 + 25x² + 300x + 900

Combine like terms:

x² + 4x² + 25x² = 30x²

120x + 300x = 420x

900 + 900 = 1800

Therefore, ||v||² = 30x² + 420x + 1800

Now, to find the minimum of this quadratic function in x. Since the coefficient of x² is positive (30), the parabola opens upwards, so the minimum occurs at the vertex.

The x-coordinate of the vertex is at -b/(2a), where a = 30, b = 420.

So, x = -420/(2*30) = -420/60 = -7.

Therefore, the minimum occurs at x = -7. Then, we can find y and z.

From equation 3: y = 2x + 30. So, substituting x = -7:

y = 2*(-7) + 30 = -14 + 30 = 16

From equation 2: z = -5x -30. Substituting x = -7:

z = -5*(-7) -30 = 35 - 30 = 5

Therefore, the vector v is (-7, 16, 5). Let me check if this satisfies the original cross product equation.

Let me compute (1, 2, -5) × (-7, 16, 5):

First component: 2*5 - (-5)*16 = 10 + 80 = 90

Second component: - [1*5 - (-5)*(-7)] = - [5 - 35] = - (-30) = 30

Third component: 1*16 - 2*(-7) = 16 + 14 = 30

Yes! That gives (90, 30, 30), which matches. So that's correct.

Now, let me check the magnitude of this vector. ||v|| = sqrt((-7)^2 + 16^2 + 5^2) = sqrt(49 + 256 + 25) = sqrt(330). Let me compute that. 330 is 10*33, so sqrt(330) is approximately 18.16... But maybe there's a way to check if this is indeed the minimal.

Alternatively, since we found the minimum by completing the square or using calculus, and since the quadratic was correctly derived, this should be the minimal magnitude. However, let me just verify that there isn't a mistake in my calculation.

Wait, let me check the calculation of ||v||² again. When x = -7, then:

x² = (-7)^2 = 49

y = 16, so y² = 256

z = 5, so z² = 25

Total: 49 + 256 + 25 = 330. Yes, that's correct. So ||v|| = sqrt(330). Is there a way to get a smaller magnitude?

Alternatively, maybe I can think of this problem in terms of vector projections. Since the cross product a × v = b, then v can be expressed as some vector. Wait, but maybe there's another approach.

Alternatively, since cross product a × v = b, then v can be found by solving the equation. But since there are infinitely many solutions, they form a line. The minimal magnitude vector would be the one that is orthogonal to the direction of the line. Wait, but how?

Alternatively, since the cross product a × v = b, then v must satisfy that b is perpendicular to a. Wait, is that true? Let me check. Since a × v is perpendicular to a, so b must be perpendicular to a. Let me check if that's the case here.

Given a = (1, 2, -5) and b = (90, 30, 30). Let me compute their dot product:

1*90 + 2*30 + (-5)*30 = 90 + 60 - 150 = 0. Yes! So, the dot product is zero, which means that b is indeed perpendicular to a. Therefore, the equation a × v = b is consistent.

Now, in general, the solutions to a × v = b are given by v = (b × a)/||a||² * a + t*a, where t is a scalar? Wait, maybe not. Wait, let me recall.

If we have a × v = b, then v can be expressed as v = (b × a)/||a||² * a + c, where c is any vector parallel to a? Wait, maybe. Let me think.

Alternatively, since a × v = b, then v must be in the plane perpendicular to b? Wait, no. Wait, the cross product a × v is perpendicular to both a and v. So, since b is perpendicular to a, then v must lie in the plane spanned by a and some other vector? Wait, maybe I need to think in terms of solving the equation.

Alternatively, since we have a system of equations, and we found that the solutions are parameterized by x, and we found the minimal magnitude by minimizing the quadratic. So, since that approach worked and gave us a valid answer, maybe that's sufficient.

But just to make sure, let me think of another approach. Let me recall that the cross product a × v = b implies that v is in the plane spanned by a and b? Wait, no. Wait, if a × v = b, then v is in the plane that is perpendicular to b? Wait, not sure.

Alternatively, since a × v = b, then v can be written as v = (b × a)/||a||² * a + k, where k is a vector such that a × k = 0, meaning k is parallel to a. Wait, let me check.

Let me recall that if a × v = b, then v can be written as v = (b × a)/||a||² * a + c, where c is any vector such that a × c = 0. Because if you take a × ( (b × a)/||a||² * a + c ) = a × ( (b × a)/||a||² * a ) + a × c. Since a × c = 0, then the first term is a × ( (b × a)/||a||² * a ). Let me compute that.

Let me compute a × ( (b × a)/||a||² * a ). Let me denote scalar multiple: Let me write it as [ (a × (b × a)) / ||a||² ] × a? Wait, no. Wait, the cross product is linear, so:

a × [ (b × a)/||a||² * a ] = [ (a × (b × a)) / ||a||² ] * a ?

Wait, no. Wait, cross product is not associative. Let me think again. Let me denote scalar k = (b × a)/||a||². Then, the term is a × (k * a) = k * (a × a) = 0, since cross product of a vector with itself is zero. Therefore, the entire expression becomes 0 + a × c = 0, since c is parallel to a. Therefore, this suggests that v = (b × a)/||a||² * a + c, but then a × v = b. Wait, but if I take v = (b × a)/||a||² * a + c, then a × v = a × ( (b × a)/||a||² * a ) + a × c = 0 + 0 = 0? That can't be. Wait, maybe my approach is wrong here.

Alternatively, perhaps I need to recall that if a × v = b, then v can be expressed as v = (b × a)/||a||² * a + t*a, where t is a scalar? Wait, let me check. Let me compute a × v where v = (b × a)/||a||² * a + t*a.

Then, a × v = a × [ (b × a)/||a||² * a + t*a ] = a × [ (b × a)/||a||² * a ] + a × (t*a) = 0 + 0 = 0. Which is not equal to b. So that approach is not working. Maybe I need to think differently.

Alternatively, perhaps the general solution is v = (b × a)/||a||² + t*a, where t is a scalar? Let me check. Let me compute a × v where v = (b × a)/||a||² + t*a.

Then, a × v = a × (b × a)/||a||² + a × (t*a) = [a × (b × a)] / ||a||² + 0.

Now, using the vector triple product identity: a × (b × a) = b(a · a) - a(a · b). Since a · b = 0 (as we saw earlier), this simplifies to b||a||² - 0 = b||a||². Therefore, [a × (b × a)] / ||a||² = b. Therefore, a × v = b. Therefore, the general solution is v = (b × a)/||a||² + t*a, where t is any scalar. Therefore, the solutions form a line in the direction of a, with a particular solution (b × a)/||a||².

Therefore, the set of solutions is all vectors of the form v = (b × a)/||a||² + t*a. Therefore, to find the vector with minimal magnitude, we can think of this as a line in the direction of a, and we need to find the point on this line closest to the origin.

Therefore, the minimal magnitude vector is the projection of the particular solution (b × a)/||a||² onto the line, or something like that. Wait, but since the line is parametrized as v = (b × a)/||a||² + t*a, then the minimal magnitude occurs when the vector v is orthogonal to the direction of the line, i.e., when the vector from the origin to v is orthogonal to the direction vector a. Wait, but how?

Alternatively, since the line is parametrized as v = c + t*a, where c = (b × a)/||a||², then the minimal ||v|| occurs when v is the projection of c onto the line? Wait, no. Wait, the minimal ||v|| occurs when the vector v is the closest point on the line to the origin. The closest point on the line to the origin is given by the projection of the origin onto the line. Wait, but since the line is v = c + t*a, then the closest point is c - (c · a)/||a||² * a. Wait, maybe?

Wait, more formally, to find the minimum of ||v|| where v = c + t*a, we can take derivative with respect to t and set to zero. Let me do that.

Let me denote v(t) = c + t*a. Then, ||v(t)||² = ||c||² + 2t*(c · a) + t²||a||². Taking derivative with respect to t: 2*(c · a) + 2t||a||². Setting to zero: 2*(c · a) + 2t||a||² = 0 → t = - (c · a)/||a||².

Therefore, the minimal ||v|| occurs at t = - (c · a)/||a||². Therefore, the minimal vector is v_min = c - (c · a)/||a||² * a.

But in our case, c = (b × a)/||a||². Therefore, let me compute c · a.

c · a = [ (b × a)/||a||² ] · a = [ (b × a) · a ] / ||a||². But (b × a) · a = 0, since the cross product is perpendicular to a. Therefore, c · a = 0. Therefore, t = -0 / ||a||² = 0. Therefore, the minimal ||v|| occurs at t = 0, so v_min = c = (b × a)/||a||².

Wait, so according to this, the minimal magnitude vector is (b × a)/||a||². But in our previous approach, we found a different vector. Let me check if these are the same.

Wait, let me compute (b × a)/||a||². First, compute b × a. Since cross product is anti-commutative, b × a = - (a × b). So, if we compute a × b, which is the original cross product given as (90, 30, 30). Wait, no. Wait, the original equation is a × v = b. So, if we compute b × a, that would be - (a × b). But in our case, a × v = b, so if we take v = (b × a)/||a||², then a × v = a × [ (b × a)/||a||² ] = [a × (b × a)] / ||a||² = [b(a · a) - a(a · b)] / ||a||². Since a · b = 0, this becomes [b||a||²]/||a||² = b. Therefore, this works. Therefore, the particular solution is v = (b × a)/||a||². However, according to the previous calculation, since c · a = 0, then the minimal vector is c, which is (b × a)/||a||². Therefore, according to this, the minimal vector is (b × a)/||a||². But in our previous approach, we found v = (-7, 16, 5). Let me check if these are the same.

First, compute b × a. Given that a = (1, 2, -5), b = (90, 30, 30). So, b × a is:

|i   j   k |
|90  30  30|
|1    2  -5|

Calculating determinant:

i*(30*(-5) - 30*2) - j*(90*(-5) - 30*1) + k*(90*2 - 30*1)

Compute each component:

First component: i*(-150 - 60) = i*(-210)

Second component: -j*(-450 - 30) = -j*(-480) = j*480

Third component: k*(180 - 30) = k*150

Therefore, b × a = (-210, 480, 150)

Then, ||a||² is 1² + 2² + (-5)² = 1 + 4 + 25 = 30.

Therefore, (b × a)/||a||² = (-210/30, 480/30, 150/30) = (-7, 16, 5). Which is exactly the vector we found earlier! So, that's the same vector. Therefore, both methods agree. Therefore, the minimal magnitude vector is (-7, 16, 5), with magnitude sqrt(330). Therefore, that's the answer.

But let me just check if there's another way to see this. Since the minimal vector is the projection of the particular solution onto the line? Wait, but in this case, since c · a = 0, the particular solution is already orthogonal to a, so it's the closest point on the line to the origin. Therefore, that's why the minimal magnitude is achieved at that particular solution. Therefore, this confirms that our answer is correct.

Alternatively, if I didn't remember the formula for the general solution, solving the system of equations and minimizing the magnitude was a valid approach, and it led me to the same answer. Therefore, I can be confident that (-7, 16, 5) is the correct vector with minimal magnitude.

Just to recap, the steps were:

1. Set up the cross product equation and derive the system of equations.

2. Solve the system, finding that the solutions are parameterized by x.

3. Express the magnitude squared as a quadratic in x, find its minimum using vertex formula.

4. Verify the result using another method involving cross product properties and vector projections.

Both methods led to the same vector, so I think that's solid.

Therefore, the answer is the vector (-7, 16, 5), which has magnitude sqrt(330). Since the problem asks for the vector with the smallest magnitude, and we have found it, I think that's the answer.

**Final Answer**
The vector $\mathbf{v}$ with the smallest magnitude is $\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}$.

To solve the problem, we are given the equation:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \mathbf{v} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}
$$

We are to find the vector $\mathbf{v}$ that satisfies this cross product and has the **smallest magnitude**.

---

### Step 1: Express the Cross Product

Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$. Then, the cross product is:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix}
2z + 5y \\
-5x - z \\
y - 2x
\end{pmatrix}
$$

Set this equal to $\begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}$, giving the system of equations:

$$
\begin{cases}
2z + 5y = 90 \quad \text{(1)} \\
-5x - z = 30 \quad \text{(2)} \\
y - 2x = 30 \quad \text{(3)}
\end{cases}
$$

---

### Step 2: Solve the System

From equation (3):
$$
y = 2x + 30
$$

Substitute into equation (1):
$$
2z + 5(2x + 30) = 90 \Rightarrow 2z + 10x + 150 = 90 \Rightarrow 2z = -10x - 60 \Rightarrow z = -5x - 30
$$

Now we have:
- $y = 2x + 30$
- $z = -5x - 30$

So, the vector $\mathbf{v}$ is parameterized as:
$$
\mathbf{v} = \begin{pmatrix} x \\ 2x + 30 \\ -5x - 30 \end{pmatrix}
$$

---

### Step 3: Minimize the Magnitude

We want to minimize the magnitude of $\mathbf{v}$, which is:

$$
\|\mathbf{v}\|^2 = x^2 + (2x + 30)^2 + (-5x - 30)^2
$$

Compute each term:

- $x^2$
- $(2x + 30)^2 = 4x^2 + 120x + 900$
- $(-5x - 30)^2 = 25x^2 + 300x + 900$

Add them:

$$
\|\mathbf{v}\|^2 = x^2 + 4x^2 + 120x + 900 + 25x^2 + 300x + 900 = 30x^2 + 420x + 1800
$$

This is a quadratic in $x$, and since the coefficient of $x^2$ is positive, the minimum occurs at the vertex:

$$
x = -\frac{b}{2a} = -\frac{420}{2 \cdot 30} = -7
$$

Substitute $x = -7$ into the expressions for $y$ and $z$:

- $y = 2(-7) + 30 = 16$
- $z = -5(-7) - 30 = 35 - 30 = 5$

So, the vector is:

$$
\mathbf{v} = \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}
$$

---

### Step 4: Verify the Cross Product

Compute:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix} =
\begin{pmatrix}
2(5) - (-5)(16) = 10 + 80 = 90 \\
-1(5) - (-5)(-7) = -5 - 35 = -40 \Rightarrow -(-40) = 40 \text{ (Wait, sign error)}
\end{pmatrix}
$$

Wait — let's recompute the cross product correctly:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix} =
\begin{pmatrix}
2(5) - (-5)(16) = 10 + 80 = 90 \\
-1(5) - (-5)(-7) = -5 - 35 = -40 \Rightarrow -(-40) = 40 \text{ (Wait, sign error)}
\end{pmatrix}
$$

Actually, the second component is:

$$
- \left[1(5) - (-5)(-7)\right] = -[5 - 35] = -(-30) = 30
$$

Third component:

$$
1(16) - 2(-7) = 16 + 14 = 30
$$

So the result is:

$$
\begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}
$$

✅ Correct.

---

### Final Answer

$$
\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, so I need to find a vector v such that when I take the cross product of the vector (1, 2, -5) with v, I get the vector (90, 30, 30). And among all possible such vectors v, I need to find the one with the smallest magnitude. Hmm, cross products... Let me recall how cross products work.

First, the cross product of two vectors in three dimensions results in a vector that's perpendicular to both of the original vectors. So, if I have vectors a and b, then a × b is perpendicular to both a and b. That means that the cross product here, which is (1, 2, -5) × v = (90, 30, 30), must be perpendicular to both (1, 2, -5) and v. Wait, but maybe that's not directly helpful here. Let me think.

Alternatively, maybe I can set up equations. Let me denote vector v as (x, y, z). Then, the cross product of (1, 2, -5) and (x, y, z) is given by the determinant of the following matrix:

|i   j   k |
|1    2  -5|
|x    y   z|

Calculating that determinant, the cross product should be:

i*(2*z - (-5)*y) - j*(1*z - (-5)*x) + k*(1*y - 2*x)

Simplifying each component:

First component (i): 2z + 5y

Second component (j): -(z + 5x) → -z -5x

Third component (k): y - 2x

So, the cross product is (2z + 5y, -z -5x, y - 2x). According to the problem, this equals (90, 30, 30). Therefore, we can set up the following system of equations:

1) 2z + 5y = 90

2) -z -5x = 30

3) y - 2x = 30

So, three equations with three variables x, y, z. Let me write them again:

Equation 1: 5y + 2z = 90

Equation 2: -5x - z = 30

Equation 3: -2x + y = 30

So, I need to solve this system. Let me try solving equations 2 and 3 first for x and y, and then substitute into equation 1 to find z.

From equation 3: y = 2x + 30. Let me write that down.

Equation 3: y = 2x + 30

Now, substitute this into equation 2: -5x - z = 30. Wait, equation 2 is -5x - z = 30. So, if I can express z in terms of x, maybe?

Let me rearrange equation 2: -z = 30 + 5x → z = -30 -5x.

So z = -5x -30.

Now, substitute y and z into equation 1: 5y + 2z = 90.

Substituting y = 2x + 30 and z = -5x -30:

5*(2x + 30) + 2*(-5x -30) = 90

Let me compute that:

First term: 5*(2x + 30) = 10x + 150

Second term: 2*(-5x -30) = -10x -60

Adding them together: 10x + 150 -10x -60 = (10x -10x) + (150 -60) = 0x + 90 = 90

So, 90 = 90. Wait, that's an identity. So, that means that equations 1, 2, 3 are dependent? So, there are infinitely many solutions? Which matches the problem statement that there are infinitely many vectors v satisfying the equation. Therefore, we need to find the vector v with the smallest magnitude.

So, since equations 2 and 3 give us expressions for y and z in terms of x, we can write v as (x, y, z) = (x, 2x + 30, -5x -30). Therefore, the vector v is parameterized by x. So, to find the vector with the smallest magnitude, we need to minimize the magnitude of v as a function of x.

So, let me write the magnitude squared of v, since minimizing the magnitude is equivalent to minimizing the magnitude squared, which is easier.

Let me denote ||v||² = x² + y² + z².

Substituting y and z:

||v||² = x² + (2x + 30)² + (-5x -30)²

Let me expand each term:

First term: x²

Second term: (2x + 30)² = 4x² + 120x + 900

Third term: (-5x -30)² = (5x + 30)² = 25x² + 300x + 900

Adding them all together:

x² + 4x² + 120x + 900 + 25x² + 300x + 900

Combine like terms:

x² + 4x² + 25x² = 30x²

120x + 300x = 420x

900 + 900 = 1800

Therefore, ||v||² = 30x² + 420x + 1800

Now, to find the minimum of this quadratic function in x. Since the coefficient of x² is positive (30), the parabola opens upwards, so the minimum occurs at the vertex.

The x-coordinate of the vertex is at -b/(2a), where a = 30, b = 420.

So, x = -420/(2*30) = -420/60 = -7.

Therefore, the minimum occurs at x = -7. Then, we can find y and z.

From equation 3: y = 2x + 30. So, substituting x = -7:

y = 2*(-7) + 30 = -14 + 30 = 16

From equation 2: z = -5x -30. Substituting x = -7:

z = -5*(-7) -30 = 35 - 30 = 5

Therefore, the vector v is (-7, 16, 5). Let me check if this satisfies the original cross product equation.

Let me compute (1, 2, -5) × (-7, 16, 5):

First component: 2*5 - (-5)*16 = 10 + 80 = 90

Second component: - [1*5 - (-5)*(-7)] = - [5 - 35] = - (-30) = 30

Third component: 1*16 - 2*(-7) = 16 + 14 = 30

Yes! That gives (90, 30, 30), which matches. So that's correct.

Now, let me check the magnitude of this vector. ||v|| = sqrt((-7)^2 + 16^2 + 5^2) = sqrt(49 + 256 + 25) = sqrt(330). Let me compute that. 330 is 10*33, so sqrt(330) is approximately 18.16... But maybe there's a way to check if this is indeed the minimal.

Alternatively, since we found the minimum by completing the square or using calculus, and since the quadratic was correctly derived, this should be the minimal magnitude. However, let me just verify that there isn't a mistake in my calculation.

Wait, let me check the calculation of ||v||² again. When x = -7, then:

x² = (-7)^2 = 49

y = 16, so y² = 256

z = 5, so z² = 25

Total: 49 + 256 + 25 = 330. Yes, that's correct. So ||v|| = sqrt(330). Is there a way to get a smaller magnitude?

Alternatively, maybe I can think of this problem in terms of vector projections. Since the cross product a × v = b, then v can be expressed as some vector. Wait, but maybe there's another approach.

Alternatively, since cross product a × v = b, then v can be found by solving the equation. But since there are infinitely many solutions, they form a line. The minimal magnitude vector would be the one that is orthogonal to the direction of the line. Wait, but how?

Alternatively, since the cross product a × v = b, then v must satisfy that b is perpendicular to a. Wait, is that true? Let me check. Since a × v is perpendicular to a, so b must be perpendicular to a. Let me check if that's the case here.

Given a = (1, 2, -5) and b = (90, 30, 30). Let me compute their dot product:

1*90 + 2*30 + (-5)*30 = 90 + 60 - 150 = 0. Yes! So, the dot product is zero, which means that b is indeed perpendicular to a. Therefore, the equation a × v = b is consistent.

Now, in general, the solutions to a × v = b are given by v = (b × a)/||a||² * a + t*a, where t is a scalar? Wait, maybe not. Wait, let me recall.

If we have a × v = b, then v can be expressed as v = (b × a)/||a||² * a + c, where c is any vector parallel to a? Wait, maybe. Let me think.

Alternatively, since a × v = b, then v must be in the plane perpendicular to b? Wait, no. Wait, the cross product a × v is perpendicular to both a and v. So, since b is perpendicular to a, then v must lie in the plane spanned by a and some other vector? Wait, maybe I need to think in terms of solving the equation.

Alternatively, since we have a system of equations, and we found that the solutions are parameterized by x, and we found the minimal magnitude by minimizing the quadratic. So, since that approach worked and gave us a valid answer, maybe that's sufficient.

But just to make sure, let me think of another approach. Let me recall that the cross product a × v = b implies that v is in the plane spanned by a and b? Wait, no. Wait, if a × v = b, then v is in the plane that is perpendicular to b? Wait, not sure.

Alternatively, since a × v = b, then v can be written as v = (b × a)/||a||² * a + k, where k is a vector such that a × k = 0, meaning k is parallel to a. Wait, let me check.

Let me recall that if a × v = b, then v can be written as v = (b × a)/||a||² * a + c, where c is any vector such that a × c = 0. Because if you take a × ( (b × a)/||a||² * a + c ) = a × ( (b × a)/||a||² * a ) + a × c. Since a × c = 0, then the first term is a × ( (b × a)/||a||² * a ). Let me compute that.

Let me compute a × ( (b × a)/||a||² * a ). Let me denote scalar multiple: Let me write it as [ (a × (b × a)) / ||a||² ] × a? Wait, no. Wait, the cross product is linear, so:

a × [ (b × a)/||a||² * a ] = [ (a × (b × a)) / ||a||² ] * a ?

Wait, no. Wait, cross product is not associative. Let me think again. Let me denote scalar k = (b × a)/||a||². Then, the term is a × (k * a) = k * (a × a) = 0, since cross product of a vector with itself is zero. Therefore, the entire expression becomes 0 + a × c = 0, since c is parallel to a. Therefore, this suggests that v = (b × a)/||a||² * a + c, but then a × v = b. Wait, but if I take v = (b × a)/||a||² * a + c, then a × v = a × ( (b × a)/||a||² * a ) + a × c = 0 + 0 = 0? That can't be. Wait, maybe my approach is wrong here.

Alternatively, perhaps I need to recall that if a × v = b, then v can be expressed as v = (b × a)/||a||² * a + t*a, where t is a scalar? Wait, let me check. Let me compute a × v where v = (b × a)/||a||² * a + t*a.

Then, a × v = a × [ (b × a)/||a||² * a + t*a ] = a × [ (b × a)/||a||² * a ] + a × (t*a) = 0 + 0 = 0. Which is not equal to b. So that approach is not working. Maybe I need to think differently.

Alternatively, perhaps the general solution is v = (b × a)/||a||² + t*a, where t is a scalar? Let me check. Let me compute a × v where v = (b × a)/||a||² + t*a.

Then, a × v = a × (b × a)/||a||² + a × (t*a) = [a × (b × a)] / ||a||² + 0.

Now, using the vector triple product identity: a × (b × a) = b(a · a) - a(a · b). Since a · b = 0 (as we saw earlier), this simplifies to b||a||² - 0 = b||a||². Therefore, [a × (b × a)] / ||a||² = b. Therefore, a × v = b. Therefore, the general solution is v = (b × a)/||a||² + t*a, where t is any scalar. Therefore, the solutions form a line in the direction of a, with a particular solution (b × a)/||a||².

Therefore, the set of solutions is all vectors of the form v = (b × a)/||a||² + t*a. Therefore, to find the vector with minimal magnitude, we can think of this as a line in the direction of a, and we need to find the point on this line closest to the origin.

Therefore, the minimal magnitude vector is the projection of the particular solution (b × a)/||a||² onto the line, or something like that. Wait, but since the line is parametrized as v = (b × a)/||a||² + t*a, then the minimal magnitude occurs when the vector v is orthogonal to the direction of the line, i.e., when the vector from the origin to v is orthogonal to the direction vector a. Wait, but how?

Alternatively, since the line is parametrized as v = c + t*a, where c = (b × a)/||a||², then the minimal ||v|| occurs when v is the projection of c onto the line? Wait, no. Wait, the minimal ||v|| occurs when the vector v is the closest point on the line to the origin. The closest point on the line to the origin is given by the projection of the origin onto the line. Wait, but since the line is v = c + t*a, then the closest point is c - (c · a)/||a||² * a. Wait, maybe?

Wait, more formally, to find the minimum of ||v|| where v = c + t*a, we can take derivative with respect to t and set to zero. Let me do that.

Let me denote v(t) = c + t*a. Then, ||v(t)||² = ||c||² + 2t*(c · a) + t²||a||². Taking derivative with respect to t: 2*(c · a) + 2t||a||². Setting to zero: 2*(c · a) + 2t||a||² = 0 → t = - (c · a)/||a||².

Therefore, the minimal ||v|| occurs at t = - (c · a)/||a||². Therefore, the minimal vector is v_min = c - (c · a)/||a||² * a.

But in our case, c = (b × a)/||a||². Therefore, let me compute c · a.

c · a = [ (b × a)/||a||² ] · a = [ (b × a) · a ] / ||a||². But (b × a) · a = 0, since the cross product is perpendicular to a. Therefore, c · a = 0. Therefore, t = -0 / ||a||² = 0. Therefore, the minimal ||v|| occurs at t = 0, so v_min = c = (b × a)/||a||².

Wait, so according to this, the minimal magnitude vector is (b × a)/||a||². But in our previous approach, we found a different vector. Let me check if these are the same.

Wait, let me compute (b × a)/||a||². First, compute b × a. Since cross product is anti-commutative, b × a = - (a × b). So, if we compute a × b, which is the original cross product given as (90, 30, 30). Wait, no. Wait, the original equation is a × v = b. So, if we compute b × a, that would be - (a × b). But in our case, a × v = b, so if we take v = (b × a)/||a||², then a × v = a × [ (b × a)/||a||² ] = [a × (b × a)] / ||a||² = [b(a · a) - a(a · b)] / ||a||². Since a · b = 0, this becomes [b||a||²]/||a||² = b. Therefore, this works. Therefore, the particular solution is v = (b × a)/||a||². However, according to the previous calculation, since c · a = 0, then the minimal vector is c, which is (b × a)/||a||². Therefore, according to this, the minimal vector is (b × a)/||a||². But in our previous approach, we found v = (-7, 16, 5). Let me check if these are the same.

First, compute b × a. Given that a = (1, 2, -5), b = (90, 30, 30). So, b × a is:

|i   j   k |
|90  30  30|
|1    2  -5|

Calculating determinant:

i*(30*(-5) - 30*2) - j*(90*(-5) - 30*1) + k*(90*2 - 30*1)

Compute each component:

First component: i*(-150 - 60) = i*(-210)

Second component: -j*(-450 - 30) = -j*(-480) = j*480

Third component: k*(180 - 30) = k*150

Therefore, b × a = (-210, 480, 150)

Then, ||a||² is 1² + 2² + (-5)² = 1 + 4 + 25 = 30.

Therefore, (b × a)/||a||² = (-210/30, 480/30, 150/30) = (-7, 16, 5). Which is exactly the vector we found earlier! So, that's the same vector. Therefore, both methods agree. Therefore, the minimal magnitude vector is (-7, 16, 5), with magnitude sqrt(330). Therefore, that's the answer.

But let me just check if there's another way to see this. Since the minimal vector is the projection of the particular solution onto the line? Wait, but in this case, since c · a = 0, the particular solution is already orthogonal to a, so it's the closest point on the line to the origin. Therefore, that's why the minimal magnitude is achieved at that particular solution. Therefore, this confirms that our answer is correct.

Alternatively, if I didn't remember the formula for the general solution, solving the system of equations and minimizing the magnitude was a valid approach, and it led me to the same answer. Therefore, I can be confident that (-7, 16, 5) is the correct vector with minimal magnitude.

Just to recap, the steps were:

1. Set up the cross product equation and derive the system of equations.

2. Solve the system, finding that the solutions are parameterized by x.

3. Express the magnitude squared as a quadratic in x, find its minimum using vertex formula.

4. Verify the result using another method involving cross product properties and vector projections.

Both methods led to the same vector, so I think that's solid.

Therefore, the answer is the vector (-7, 16, 5), which has magnitude sqrt(330). Since the problem asks for the vector with the smallest magnitude, and we have found it, I think that's the answer.

**Final Answer**
The vector $\mathbf{v}$ with the smallest magnitude is $\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}$.

To solve the problem, we are given the equation:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \mathbf{v} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}
$$

We are to find the vector $\mathbf{v}$ that satisfies this cross product and has the **smallest magnitude**.

---

### Step 1: Express the Cross Product

Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$. Then, the cross product is:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix}
2z + 5y \\
-5x - z \\
y - 2x
\end{pmatrix}
$$

Set this equal to $\begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}$, giving the system of equations:

$$
\begin{cases}
2z + 5y = 90 \quad \text{(1)} \\
-5x - z = 30 \quad \text{(2)} \\
y - 2x = 30 \quad \text{(3)}
\end{cases}
$$

---

### Step 2: Solve the System

From equation (3):
$$
y = 2x + 30
$$

Substitute into equation (1):
$$
2z + 5(2x + 30) = 90 \Rightarrow 2z + 10x + 150 = 90 \Rightarrow 2z = -10x - 60 \Rightarrow z = -5x - 30
$$

Now we have:
- $y = 2x + 30$
- $z = -5x - 30$

So, the vector $\mathbf{v}$ is parameterized as:
$$
\mathbf{v} = \begin{pmatrix} x \\ 2x + 30 \\ -5x - 30 \end{pmatrix}
$$

---

### Step 3: Minimize the Magnitude

We want to minimize the magnitude of $\mathbf{v}$, which is:

$$
\|\mathbf{v}\|^2 = x^2 + (2x + 30)^2 + (-5x - 30)^2
$$

Compute each term:

- $x^2$
- $(2x + 30)^2 = 4x^2 + 120x + 900$
- $(-5x - 30)^2 = 25x^2 + 300x + 900$

Add them:

$$
\|\mathbf{v}\|^2 = x^2 + 4x^2 + 120x + 900 + 25x^2 + 300x + 900 = 30x^2 + 420x + 1800
$$

This is a quadratic in $x$, and since the coefficient of $x^2$ is positive, the minimum occurs at the vertex:

$$
x = -\frac{b}{2a} = -\frac{420}{2 \cdot 30} = -7
$$

Substitute $x = -7$ into the expressions for $y$ and $z$:

- $y = 2(-7) + 30 = 16$
- $z = -5(-7) - 30 = 35 - 30 = 5$

So, the vector is:

$$
\mathbf{v} = \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}
$$

---

### Step 4: Verify the Cross Product

Compute:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix} =
\begin{pmatrix}
2(5) - (-5)(16) = 10 + 80 = 90 \\
-1(5) - (-5)(-7) = -5 - 35 = -40 \Rightarrow -(-40) = 40 \text{ (Wait, sign error)}
\end{pmatrix}
$$

Wait — let's recompute the cross product correctly:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix} =
\begin{pmatrix}
2(5) - (-5)(16) = 10 + 80 = 90 \\
-1(5) - (-5)(-7) = -5 - 35 = -40 \Rightarrow -(-40) = 40 \text{ (Wait, sign error)}
\end{pmatrix}
$$

Actually, the second component is:

$$
- \left[1(5) - (-5)(-7)\right] = -[5 - 35] = -(-30) = 30
$$

Third component:

$$
1(16) - 2(-7) = 16 + 14 = 30
$$

So the result is:

$$
\begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}
$$

✅ Correct.

---

### Final Answer

$$
\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}
$$

Solution:
Let $\mathbf{v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}.$  Then
\[\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix},\]so
\[\begin{pmatrix} 5b + 2c \\ -5a - c \\ -2a + b \end{pmatrix} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}.\]Comparing the components, we get
\begin{align*}
5b + 2c &= 90, \\
-5a - c &= 30, \\
-2a + b &= 30.
\end{align*}From the second equation, $c = -5a - 30.$  From the third equation, $b = 2a + 30.$  We want to minimize the magnitude of $\mathbf{v},$ which is equivalent to minimizing
\[a^2 + b^2 + c^2 = a^2 + (2a + 30)^2 + (-5a - 30)^2 = 30a^2 + 420a + 1800.\]Completing the square, we get $30 (a + 7)^2 + 330,$ so the magnitude is minimized when $a = -7.$  Then $b = 16$ and $c = 5,$ so the vector $\mathbf{v}$ we seek is $\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.$