Explanation:
Okay, so I need to find the equation of the plane that contains the three points given: (0, -1, -1), (-4, 4, 4), and (4, 5, 1). The answer should be in the form Ax + By + Cz + D = 0, with A, B, C, D as integers, A positive, and the greatest common divisor of their absolute values being 1. Alright, let me recall how to do this.

First, I remember that to find the equation of a plane, you need a point on the plane and a normal vector to the plane. Since we have three points, we can use them to find two vectors lying on the plane, then take their cross product to get the normal vector. Once we have the normal vector, we can plug it into the plane equation along with one of the points to solve for D.

Let me write down the points again for clarity:

Point P: (0, -1, -1)
Point Q: (-4, 4, 4)
Point R: (4, 5, 1)

So, first step: find two vectors on the plane. Let me subtract coordinates of the points to get vectors PQ and PR or something like that.

Let me compute vector PQ and PR. Wait, actually, vectors can be found by subtracting coordinates of two points. Let me pick vectors PQ and PR. Wait, but maybe it's better to use vectors from the same point. Let me think.

Alternatively, I can take vectors PQ and QR or PR and QR. Let me just pick two vectors. Let me take vectors from point P to Q and from point P to R. That should work.

So vector PQ is Q - P. Let me compute that:

Q - P = (-4 - 0, 4 - (-1), 4 - (-1)) = (-4, 5, 5)

Similarly, vector PR is R - P = (4 - 0, 5 - (-1), 1 - (-1)) = (4, 6, 2)

Okay, so now I have two vectors in the plane: PQ = (-4, 5, 5) and PR = (4, 6, 2). Now, the cross product of these two vectors will give me a normal vector to the plane.

Let me recall how to compute the cross product. If I have two vectors, say vector a = (a1, a2, a3) and vector b = (b1, b2, b3), then their cross product is:

(a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

Alternatively, using the determinant formula with unit vectors i, j, k.

Let me write that out for PQ and PR.

Let me denote vector PQ as a = (-4, 5, 5) and vector PR as b = (4, 6, 2). Then cross product a × b is:

|i   j   k |
|-4 5   5|
|4   6   2|

Calculating determinant:

i*(5*2 - 5*6) - j*(-4*2 - 5*4) + k*(-4*6 - 5*4)

Let me compute each component step by step.

First component (i):

5*2 - 5*6 = 10 - 30 = -20

Second component (j):

- [ (-4)*2 - 5*4 ] = - [ -8 - 20 ] = - [ -28 ] = 28

Wait, hold on. Wait, the formula is:

For the cross product, the j component is subtracted, so it's - (a1b3 - a3b1). Wait, maybe I should double-check.

Alternatively, using the standard formula:

If a = (a1, a2, a3), b = (b1, b2, b3), then cross product is:

(a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

So for the first component (x-component): a2b3 - a3b2 = 5*2 - 5*6 = 10 - 30 = -20

Second component (y-component): a3b1 - a1b3 = 5*4 - (-4)*2 = 20 - (-8) = 20 + 8 = 28

Third component (z-component): a1b2 - a2b1 = (-4)*6 - 5*4 = -24 - 20 = -44

Therefore, the cross product vector is (-20, 28, -44). So this is the normal vector n = (-20, 28, -44).

But we can simplify this vector if possible. Let me check if there is a common factor. Let's see:

The components are -20, 28, -44. Let me find the greatest common divisor of 20, 28, 44.

Factors of 20: 2, 4, 5, 10, 20

Factors of 28: 2, 4, 7, 14, 28

Factors of 44: 2, 4, 11, 22, 44

Common factors are 2 and 4. Let me check if 4 divides all:

-20 divided by 4 is -5, 28 divided by 4 is 7, -44 divided by 4 is -11. So yes, we can divide by 4. So the normal vector can be simplified to (-5, 7, -11). Let me check:

Original cross product: (-20, 28, -44). Divided by 4: (-5, 7, -11). Yes. So that's a simpler normal vector.

Alternatively, since the direction of the normal vector doesn't matter (as long as it's perpendicular), but we can also take the opposite direction if needed. However, since we need A > 0 in the plane equation, we need to make sure that the coefficient A is positive. Let me check the normal vector (-5, 7, -11). The first component is -5, which is negative. So if we take the normal vector as (-5, 7, -11), then in the plane equation, A would be -5, but we need A > 0. Therefore, we can multiply the entire normal vector by -1 to get (5, -7, 11). That way, A is positive. So the normal vector is (5, -7, 11). Let me confirm that.

Yes, because multiplying by -1 reverses the direction, but it's still a normal vector. So that's better for our purposes.

So now, the normal vector is (5, -7, 11). Now, we can use this normal vector and one of the points to write the plane equation.

The general equation of a plane is A(x - x0) + B(y - y0) + C(z - z0) = 0, where (x0, y0, z0) is a point on the plane, and (A, B, C) is the normal vector.

Let me pick point P (0, -1, -1) as the point on the plane. So plugging into the equation:

5(x - 0) + (-7)(y - (-1)) + 11(z - (-1)) = 0

Simplify:

5x -7(y + 1) + 11(z + 1) = 0

Expanding:

5x -7y -7 + 11z + 11 = 0

Combine constants: -7 + 11 = 4

So:

5x -7y + 11z + 4 = 0

Wait, so the equation is 5x -7y +11z +4 = 0. Let me check if this is correct. Let me verify if all three points satisfy this equation.

First, check point P: (0, -1, -1)

Plug in: 5*0 -7*(-1) +11*(-1) +4 = 0 +7 -11 +4 = 0. 7 -11 is -4, +4 is 0. Correct.

Check point Q: (-4, 4, 4)

Plug in: 5*(-4) -7*(4) +11*(4) +4 = -20 -28 +44 +4 = (-20 -28) + (44 +4) = -48 + 48 = 0. Correct.

Check point R: (4, 5, 1)

Plug in: 5*4 -7*5 +11*1 +4 = 20 -35 +11 +4 = (20 -35) + (11 +4) = (-15) + 15 = 0. Correct. Perfect, all three points satisfy the equation. So that seems right.

But let me check if the coefficients have a common divisor. The coefficients are 5, -7, 11, and 4. Let me check gcd of 5, 7, 11, 4. Well, 5 is prime, 7 is prime, 11 is prime, and 4 is 2^2. The gcd of all these numbers is 1, since they don't share any common factors besides 1. Therefore, the equation 5x -7y +11z +4 = 0 satisfies the condition that gcd(|A|, |B|, |C|, |D|) = 1, and A is positive. So that's the answer?

Wait, but let me check again. The problem says "Enter your answer in the form Ax + By + Cz + D = 0, where A, B, C, D are integers such that A > 0 and gcd(|A|, |B|, |C|, |D|) = 1."

So our equation is 5x -7y +11z +4 = 0. Let me check the gcd of 5, 7, 11, 4. The absolute values are 5,7,11,4. The gcd of these numbers is 1, since 5 is prime, 7 is prime, 11 is prime, and 4 is 2^2. There is no common divisor other than 1. So that's good. Also, A is 5, which is positive. So that seems to satisfy all the conditions.

But let me just make sure that I didn't make any mistakes in the cross product or the calculation. Let me recheck the cross product step.

Original vectors PQ = (-4, 5, 5) and PR = (4, 6, 2). Cross product:

i*(5*2 - 5*6) - j*(-4*2 - 5*4) + k*(-4*6 - 5*4)

Wait, earlier I calculated the cross product as (-20, 28, -44). Let me check again:

First component: (5*2 - 5*6) = 10 - 30 = -20. Correct.

Second component: -( (-4*2 - 5*4) ) = - ( -8 - 20 ) = -(-28) = 28. Correct.

Third component: (-4*6 - 5*4) = -24 -20 = -44. Correct. So cross product is (-20, 28, -44). Then dividing by 4 gives (-5, 7, -11). Then multiplying by -1 gives (5, -7, 11). Correct.

Then using point P (0, -1, -1):

5(x - 0) -7(y - (-1)) + 11(z - (-1)) = 0 => 5x -7(y +1) +11(z +1) = 0 => 5x -7y -7 +11z +11 = 0 => 5x -7y +11z +4 = 0. Correct.

So that seems correct. Therefore, the answer is 5x -7y +11z +4 = 0. Let me check once more if there's a possible mistake in signs or coefficients.

Alternatively, maybe I can use another point to derive the equation. Let me try using point Q (-4, 4, 4). Let me use the normal vector (5, -7, 11). The plane equation would be 5(x +4) -7(y -4) +11(z -4) = 0. Wait, no. Wait, the general equation is A(x - x0) + B(y - y0) + C(z - z0) = 0. So if I use point Q (-4, 4, 4), then:

5(x - (-4)) -7(y - 4) +11(z - 4) = 0 => 5(x +4) -7(y -4) +11(z -4) = 0.

Expanding:

5x +20 -7y +28 +11z -44 = 0 => 5x -7y +11z + (20 +28 -44) = 0 => 5x -7y +11z +4 = 0. Same equation. Correct.

Similarly, using point R (4,5,1):

5(x -4) -7(y -5) +11(z -1) = 0 => 5x -20 -7y +35 +11z -11 = 0 => 5x -7y +11z + ( -20 +35 -11 ) = 5x -7y +11z +4 = 0. Correct. So all three points check out. Therefore, the equation is correct.

Therefore, the answer is 5x -7y +11z +4 = 0. So in the required form, that's Ax + By + Cz + D = 0 with A=5, B=-7, C=11, D=4. Since A is positive and gcd(5,7,11,4)=1, this should be the correct answer.

But just to be thorough, let me check if there's another way to approach this problem, maybe by solving equations. Let me try that method as a cross-check.

The general equation of a plane is Ax + By + Cz + D = 0. Since the three points lie on the plane, plugging them into the equation gives three equations:

For point P (0, -1, -1):

A*0 + B*(-1) + C*(-1) + D = 0 => -B - C + D = 0 => -B - C + D = 0. Equation (1)

For point Q (-4, 4, 4):

A*(-4) + B*4 + C*4 + D = 0 => -4A +4B +4C + D = 0. Equation (2)

For point R (4, 5, 1):

A*4 + B*5 + C*1 + D = 0 => 4A +5B + C + D = 0. Equation (3)

So now, we have three equations:

1) -B - C + D = 0

2) -4A +4B +4C + D = 0

3) 4A +5B + C + D = 0

We need to solve for A, B, C, D. Since there are four variables, but we can set one variable as a parameter or find ratios. However, since we need integer solutions with gcd 1, perhaps we can solve the system.

Let me write equations (1), (2), (3):

Equation (1): -B - C + D = 0 => D = B + C

Equation (2): -4A +4B +4C + D = 0

Equation (3): 4A +5B + C + D = 0

Now, substitute D from equation (1) into equations (2) and (3):

Equation (2): -4A +4B +4C + (B + C) = 0 => -4A +4B +4C + B + C = 0 => -4A +5B +5C = 0 => -4A +5B +5C = 0. Let me call this equation (2a)

Equation (3): 4A +5B + C + (B + C) = 0 => 4A +5B + C + B + C = 0 => 4A +6B + 2C = 0. Let me call this equation (3a)

So now, equations (2a) and (3a):

(2a): -4A +5B +5C = 0

(3a): 4A +6B +2C = 0

Let me add equations (2a) and (3a) together to eliminate A:

(-4A +4A) + (5B +6B) + (5C +2C) = 0 + 0 => 0 + 11B +7C = 0 => 11B +7C = 0. Let me call this equation (4)

So equation (4): 11B +7C = 0 => 11B = -7C => B = (-7/11)C

Hmm, so B is proportional to C. Let me express B in terms of C. Let me let C = 11k, then B = -7k, where k is some constant. Let me choose k as integer to make B and C integers.

So let me set C = 11k, B = -7k. Then, from equation (1), D = B + C = -7k +11k = 4k.

Now, let me substitute B and C into equation (2a) or (3a) to find A. Let me use equation (2a):

-4A +5B +5C = 0 => -4A +5*(-7k) +5*(11k) = 0 => -4A -35k +55k = 0 => -4A +20k = 0 => -4A = -20k => A = 5k.

So A = 5k, B = -7k, C = 11k, D = 4k.

Therefore, the plane equation is:

A x + B y + C z + D = 0 => 5k x -7k y +11k z +4k = 0.

We can factor out k:

k*(5x -7y +11z +4) = 0.

Since k ≠ 0 (otherwise all coefficients would be zero, which is not a plane), we can divide both sides by k, getting:

5x -7y +11z +4 = 0.

Which is exactly the same equation as before! So this confirms that the equation is correct. Therefore, the answer is indeed 5x -7y +11z +4 = 0.

Now, checking the gcd condition. The coefficients are 5, -7, 11, 4. The absolute values are 5,7,11,4. The gcd of these numbers is 1, as 5,7,11 are primes and 4 is 2², so no common divisors other than 1. Therefore, this satisfies the condition. Also, A = 5 > 0. Perfect.

Therefore, the answer is 5x -7y +11z +4 = 0.

**Final Answer**
The equation of the plane is \boxed{5x - 7y + 11z + 4 = 0}.

To find the equation of the plane containing the points $ P(0, -1, -1) $, $ Q(-4, 4, 4) $, and $ R(4, 5, 1) $, we proceed as follows:

---

### **Step 1: Find Two Vectors in the Plane**

We compute two vectors that lie on the plane by subtracting coordinates of the given points:

- Vector $ \vec{PQ} = Q - P = (-4 - 0, 4 - (-1), 4 - (-1)) = (-4, 5, 5) $
- Vector $ \vec{PR} = R - P = (4 - 0, 5 - (-1), 1 - (-1)) = (4, 6, 2) $

---

### **Step 2: Compute the Normal Vector via Cross Product**

The cross product of $ \vec{PQ} $ and $ \vec{PR} $ gives a vector normal to the plane:

$$
\vec{n} = \vec{PQ} \times \vec{PR} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-4 & 5 & 5 \\
4 & 6 & 2
\end{vmatrix}
= \mathbf{i}(5 \cdot 2 - 5 \cdot 6) - \mathbf{j}(-4 \cdot 2 - 5 \cdot 4) + \mathbf{k}(-4 \cdot 6 - 5 \cdot 4)
$$

$$
= \mathbf{i}(-20) - \mathbf{j}(-28) + \mathbf{k}(-44) = (-20, 28, -44)
$$

We simplify this vector by dividing by the greatest common divisor of its components (which is 4):

$$
\vec{n} = (-5, 7, -11)
$$

To ensure the coefficient $ A > 0 $, we multiply the vector by $-1$:

$$
\vec{n} = (5, -7, 11)
$$

---

### **Step 3: Use the Point-Normal Form of the Plane Equation**

Using point $ P(0, -1, -1) $ and the normal vector $ (5, -7, 11) $, the plane equation is:

$$
5(x - 0) - 7(y + 1) + 11(z + 1) = 0
$$

Simplify:

$$
5x - 7y - 7 + 11z + 11 = 0 \Rightarrow 5x - 7y + 11z + 4 = 0
$$

---

### **Step 4: Verify the Equation**

We verify that all three points satisfy the equation:

- For $ P(0, -1, -1) $: $ 5(0) - 7(-1) + 11(-1) + 4 = 0 + 7 - 11 + 4 = 0 $
- For $ Q(-4, 4, 4) $: $ 5(-4) - 7(4) + 11(4) + 4 = -20 - 28 + 44 + 4 = 0 $
- For $ R(4, 5, 1) $: $ 5(4) - 7(5) + 11(1) + 4 = 20 - 35 + 11 + 4 = 0 $

All points satisfy the equation.

---

### **Step 5: Check the Final Conditions**

- The equation is in the form $ Ax + By + Cz + D = 0 $, with $ A = 5 > 0 $
- The coefficients are $ A = 5 $, $ B = -7 $, $ C = 11 $, $ D = 4 $
- The greatest common divisor of $ |5|, |7|, |11|, |4| $ is 1

---

### **Final Answer**

$$
\boxed{5x - 7y + 11z + 4 = 0}
$$

[Assistant]
Explanation:
Okay, so I need to find the equation of the plane that contains the three points given: (0, -1, -1), (-4, 4, 4), and (4, 5, 1). The answer should be in the form Ax + By + Cz + D = 0, with A, B, C, D as integers, A positive, and the greatest common divisor of their absolute values being 1. Alright, let me recall how to do this.

First, I remember that to find the equation of a plane, you need a point on the plane and a normal vector to the plane. Since we have three points, we can use them to find two vectors lying on the plane, then take their cross product to get the normal vector. Once we have the normal vector, we can plug it into the plane equation along with one of the points to solve for D.

Let me write down the points again for clarity:

Point P: (0, -1, -1)
Point Q: (-4, 4, 4)
Point R: (4, 5, 1)

So, first step: find two vectors on the plane. Let me subtract coordinates of the points to get vectors PQ and PR or something like that.

Let me compute vector PQ and PR. Wait, actually, vectors can be found by subtracting coordinates of two points. Let me pick vectors PQ and PR. Wait, but maybe it's better to use vectors from the same point. Let me think.

Alternatively, I can take vectors PQ and QR or PR and QR. Let me just pick two vectors. Let me take vectors from point P to Q and from point P to R. That should work.

So vector PQ is Q - P. Let me compute that:

Q - P = (-4 - 0, 4 - (-1), 4 - (-1)) = (-4, 5, 5)

Similarly, vector PR is R - P = (4 - 0, 5 - (-1), 1 - (-1)) = (4, 6, 2)

Okay, so now I have two vectors in the plane: PQ = (-4, 5, 5) and PR = (4, 6, 2). Now, the cross product of these two vectors will give me a normal vector to the plane.

Let me recall how to compute the cross product. If I have two vectors, say vector a = (a1, a2, a3) and vector b = (b1, b2, b3), then their cross product is:

(a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

Alternatively, using the determinant formula with unit vectors i, j, k.

Let me write that out for PQ and PR.

Let me denote vector PQ as a = (-4, 5, 5) and vector PR as b = (4, 6, 2). Then cross product a × b is:

|i   j   k |
|-4 5   5|
|4   6   2|

Calculating determinant:

i*(5*2 - 5*6) - j*(-4*2 - 5*4) + k*(-4*6 - 5*4)

Let me compute each component step by step.

First component (i):

5*2 - 5*6 = 10 - 30 = -20

Second component (j):

- [ (-4)*2 - 5*4 ] = - [ -8 - 20 ] = - [ -28 ] = 28

Wait, hold on. Wait, the formula is:

For the cross product, the j component is subtracted, so it's - (a1b3 - a3b1). Wait, maybe I should double-check.

Alternatively, using the standard formula:

If a = (a1, a2, a3), b = (b1, b2, b3), then cross product is:

(a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

So for the first component (x-component): a2b3 - a3b2 = 5*2 - 5*6 = 10 - 30 = -20

Second component (y-component): a3b1 - a1b3 = 5*4 - (-4)*2 = 20 - (-8) = 20 + 8 = 28

Third component (z-component): a1b2 - a2b1 = (-4)*6 - 5*4 = -24 - 20 = -44

Therefore, the cross product vector is (-20, 28, -44). So this is the normal vector n = (-20, 28, -44).

But we can simplify this vector if possible. Let me check if there is a common factor. Let's see:

The components are -20, 28, -44. Let me find the greatest common divisor of 20, 28, 44.

Factors of 20: 2, 4, 5, 10, 20

Factors of 28: 2, 4, 7, 14, 28

Factors of 44: 2, 4, 11, 22, 44

Common factors are 2 and 4. Let me check if 4 divides all:

-20 divided by 4 is -5, 28 divided by 4 is 7, -44 divided by 4 is -11. So yes, we can divide by 4. So the normal vector can be simplified to (-5, 7, -11). Let me check:

Original cross product: (-20, 28, -44). Divided by 4: (-5, 7, -11). Yes. So that's a simpler normal vector.

Alternatively, since the direction of the normal vector doesn't matter (as long as it's perpendicular), but we can also take the opposite direction if needed. However, since we need A > 0 in the plane equation, we need to make sure that the coefficient A is positive. Let me check the normal vector (-5, 7, -11). The first component is -5, which is negative. So if we take the normal vector as (-5, 7, -11), then in the plane equation, A would be -5, but we need A > 0. Therefore, we can multiply the entire normal vector by -1 to get (5, -7, 11). That way, A is positive. So the normal vector is (5, -7, 11). Let me confirm that.

Yes, because multiplying by -1 reverses the direction, but it's still a normal vector. So that's better for our purposes.

So now, the normal vector is (5, -7, 11). Now, we can use this normal vector and one of the points to write the plane equation.

The general equation of a plane is A(x - x0) + B(y - y0) + C(z - z0) = 0, where (x0, y0, z0) is a point on the plane, and (A, B, C) is the normal vector.

Let me pick point P (0, -1, -1) as the point on the plane. So plugging into the equation:

5(x - 0) + (-7)(y - (-1)) + 11(z - (-1)) = 0

Simplify:

5x -7(y + 1) + 11(z + 1) = 0

Expanding:

5x -7y -7 + 11z + 11 = 0

Combine constants: -7 + 11 = 4

So:

5x -7y + 11z + 4 = 0

Wait, so the equation is 5x -7y +11z +4 = 0. Let me check if this is correct. Let me verify if all three points satisfy this equation.

First, check point P: (0, -1, -1)

Plug in: 5*0 -7*(-1) +11*(-1) +4 = 0 +7 -11 +4 = 0. 7 -11 is -4, +4 is 0. Correct.

Check point Q: (-4, 4, 4)

Plug in: 5*(-4) -7*(4) +11*(4) +4 = -20 -28 +44 +4 = (-20 -28) + (44 +4) = -48 + 48 = 0. Correct.

Check point R: (4, 5, 1)

Plug in: 5*4 -7*5 +11*1 +4 = 20 -35 +11 +4 = (20 -35) + (11 +4) = (-15) + 15 = 0. Correct. Perfect, all three points satisfy the equation. So that seems right.

But let me check if the coefficients have a common divisor. The coefficients are 5, -7, 11, and 4. Let me check gcd of 5, 7, 11, 4. Well, 5 is prime, 7 is prime, 11 is prime, and 4 is 2^2. The gcd of all these numbers is 1, since they don't share any common factors besides 1. Therefore, the equation 5x -7y +11z +4 = 0 satisfies the condition that gcd(|A|, |B|, |C|, |D|) = 1, and A is positive. So that's the answer?

Wait, but let me check again. The problem says "Enter your answer in the form Ax + By + Cz + D = 0, where A, B, C, D are integers such that A > 0 and gcd(|A|, |B|, |C|, |D|) = 1."

So our equation is 5x -7y +11z +4 = 0. Let me check the gcd of 5, 7, 11, 4. The absolute values are 5,7,11,4. The gcd of these numbers is 1, since 5 is prime, 7 is prime, 11 is prime, and 4 is 2^2. There is no common divisor other than 1. So that's good. Also, A is 5, which is positive. So that seems to satisfy all the conditions.

But let me just make sure that I didn't make any mistakes in the cross product or the calculation. Let me recheck the cross product step.

Original vectors PQ = (-4, 5, 5) and PR = (4, 6, 2). Cross product:

i*(5*2 - 5*6) - j*(-4*2 - 5*4) + k*(-4*6 - 5*4)

Wait, earlier I calculated the cross product as (-20, 28, -44). Let me check again:

First component: (5*2 - 5*6) = 10 - 30 = -20. Correct.

Second component: -( (-4*2 - 5*4) ) = - ( -8 - 20 ) = -(-28) = 28. Correct.

Third component: (-4*6 - 5*4) = -24 -20 = -44. Correct. So cross product is (-20, 28, -44). Then dividing by 4 gives (-5, 7, -11). Then multiplying by -1 gives (5, -7, 11). Correct.

Then using point P (0, -1, -1):

5(x - 0) -7(y - (-1)) + 11(z - (-1)) = 0 => 5x -7(y +1) +11(z +1) = 0 => 5x -7y -7 +11z +11 = 0 => 5x -7y +11z +4 = 0. Correct.

So that seems correct. Therefore, the answer is 5x -7y +11z +4 = 0. Let me check once more if there's a possible mistake in signs or coefficients.

Alternatively, maybe I can use another point to derive the equation. Let me try using point Q (-4, 4, 4). Let me use the normal vector (5, -7, 11). The plane equation would be 5(x +4) -7(y -4) +11(z -4) = 0. Wait, no. Wait, the general equation is A(x - x0) + B(y - y0) + C(z - z0) = 0. So if I use point Q (-4, 4, 4), then:

5(x - (-4)) -7(y - 4) +11(z - 4) = 0 => 5(x +4) -7(y -4) +11(z -4) = 0.

Expanding:

5x +20 -7y +28 +11z -44 = 0 => 5x -7y +11z + (20 +28 -44) = 0 => 5x -7y +11z +4 = 0. Same equation. Correct.

Similarly, using point R (4,5,1):

5(x -4) -7(y -5) +11(z -1) = 0 => 5x -20 -7y +35 +11z -11 = 0 => 5x -7y +11z + ( -20 +35 -11 ) = 5x -7y +11z +4 = 0. Correct. So all three points check out. Therefore, the equation is correct.

Therefore, the answer is 5x -7y +11z +4 = 0. So in the required form, that's Ax + By + Cz + D = 0 with A=5, B=-7, C=11, D=4. Since A is positive and gcd(5,7,11,4)=1, this should be the correct answer.

But just to be thorough, let me check if there's another way to approach this problem, maybe by solving equations. Let me try that method as a cross-check.

The general equation of a plane is Ax + By + Cz + D = 0. Since the three points lie on the plane, plugging them into the equation gives three equations:

For point P (0, -1, -1):

A*0 + B*(-1) + C*(-1) + D = 0 => -B - C + D = 0 => -B - C + D = 0. Equation (1)

For point Q (-4, 4, 4):

A*(-4) + B*4 + C*4 + D = 0 => -4A +4B +4C + D = 0. Equation (2)

For point R (4, 5, 1):

A*4 + B*5 + C*1 + D = 0 => 4A +5B + C + D = 0. Equation (3)

So now, we have three equations:

1) -B - C + D = 0

2) -4A +4B +4C + D = 0

3) 4A +5B + C + D = 0

We need to solve for A, B, C, D. Since there are four variables, but we can set one variable as a parameter or find ratios. However, since we need integer solutions with gcd 1, perhaps we can solve the system.

Let me write equations (1), (2), (3):

Equation (1): -B - C + D = 0 => D = B + C

Equation (2): -4A +4B +4C + D = 0

Equation (3): 4A +5B + C + D = 0

Now, substitute D from equation (1) into equations (2) and (3):

Equation (2): -4A +4B +4C + (B + C) = 0 => -4A +4B +4C + B + C = 0 => -4A +5B +5C = 0 => -4A +5B +5C = 0. Let me call this equation (2a)

Equation (3): 4A +5B + C + (B + C) = 0 => 4A +5B + C + B + C = 0 => 4A +6B + 2C = 0. Let me call this equation (3a)

So now, equations (2a) and (3a):

(2a): -4A +5B +5C = 0

(3a): 4A +6B +2C = 0

Let me add equations (2a) and (3a) together to eliminate A:

(-4A +4A) + (5B +6B) + (5C +2C) = 0 + 0 => 0 + 11B +7C = 0 => 11B +7C = 0. Let me call this equation (4)

So equation (4): 11B +7C = 0 => 11B = -7C => B = (-7/11)C

Hmm, so B is proportional to C. Let me express B in terms of C. Let me let C = 11k, then B = -7k, where k is some constant. Let me choose k as integer to make B and C integers.

So let me set C = 11k, B = -7k. Then, from equation (1), D = B + C = -7k +11k = 4k.

Now, let me substitute B and C into equation (2a) or (3a) to find A. Let me use equation (2a):

-4A +5B +5C = 0 => -4A +5*(-7k) +5*(11k) = 0 => -4A -35k +55k = 0 => -4A +20k = 0 => -4A = -20k => A = 5k.

So A = 5k, B = -7k, C = 11k, D = 4k.

Therefore, the plane equation is:

A x + B y + C z + D = 0 => 5k x -7k y +11k z +4k = 0.

We can factor out k:

k*(5x -7y +11z +4) = 0.

Since k ≠ 0 (otherwise all coefficients would be zero, which is not a plane), we can divide both sides by k, getting:

5x -7y +11z +4 = 0.

Which is exactly the same equation as before! So this confirms that the equation is correct. Therefore, the answer is indeed 5x -7y +11z +4 = 0.

Now, checking the gcd condition. The coefficients are 5, -7, 11, 4. The absolute values are 5,7,11,4. The gcd of these numbers is 1, as 5,7,11 are primes and 4 is 2², so no common divisors other than 1. Therefore, this satisfies the condition. Also, A = 5 > 0. Perfect.

Therefore, the answer is 5x -7y +11z +4 = 0.

**Final Answer**
The equation of the plane is \boxed{5x - 7y + 11z + 4 = 0}.

To find the equation of the plane containing the points $ P(0, -1, -1) $, $ Q(-4, 4, 4) $, and $ R(4, 5, 1) $, we proceed as follows:

---

### **Step 1: Find Two Vectors in the Plane**

We compute two vectors that lie on the plane by subtracting coordinates of the given points:

- Vector $ \vec{PQ} = Q - P = (-4 - 0, 4 - (-1), 4 - (-1)) = (-4, 5, 5) $
- Vector $ \vec{PR} = R - P = (4 - 0, 5 - (-1), 1 - (-1)) = (4, 6, 2) $

---

### **Step 2: Compute the Normal Vector via Cross Product**

The cross product of $ \vec{PQ} $ and $ \vec{PR} $ gives a vector normal to the plane:

$$
\vec{n} = \vec{PQ} \times \vec{PR} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-4 & 5 & 5 \\
4 & 6 & 2
\end{vmatrix}
= \mathbf{i}(5 \cdot 2 - 5 \cdot 6) - \mathbf{j}(-4 \cdot 2 - 5 \cdot 4) + \mathbf{k}(-4 \cdot 6 - 5 \cdot 4)
$$

$$
= \mathbf{i}(-20) - \mathbf{j}(-28) + \mathbf{k}(-44) = (-20, 28, -44)
$$

We simplify this vector by dividing by the greatest common divisor of its components (which is 4):

$$
\vec{n} = (-5, 7, -11)
$$

To ensure the coefficient $ A > 0 $, we multiply the vector by $-1$:

$$
\vec{n} = (5, -7, 11)
$$

---

### **Step 3: Use the Point-Normal Form of the Plane Equation**

Using point $ P(0, -1, -1) $ and the normal vector $ (5, -7, 11) $, the plane equation is:

$$
5(x - 0) - 7(y + 1) + 11(z + 1) = 0
$$

Simplify:

$$
5x - 7y - 7 + 11z + 11 = 0 \Rightarrow 5x - 7y + 11z + 4 = 0
$$

---

### **Step 4: Verify the Equation**

We verify that all three points satisfy the equation:

- For $ P(0, -1, -1) $: $ 5(0) - 7(-1) + 11(-1) + 4 = 0 + 7 - 11 + 4 = 0 $
- For $ Q(-4, 4, 4) $: $ 5(-4) - 7(4) + 11(4) + 4 = -20 - 28 + 44 + 4 = 0 $
- For $ R(4, 5, 1) $: $ 5(4) - 7(5) + 11(1) + 4 = 20 - 35 + 11 + 4 = 0 $

All points satisfy the equation.

---

### **Step 5: Check the Final Conditions**

- The equation is in the form $ Ax + By + Cz + D = 0 $, with $ A = 5 > 0 $
- The coefficients are $ A = 5 $, $ B = -7 $, $ C = 11 $, $ D = 4 $
- The greatest common divisor of $ |5|, |7|, |11|, |4| $ is 1

---

### **Final Answer**

$$
\boxed{5x - 7y + 11z + 4 = 0}
$$

Solution:
Let $\mathbf{a} = \begin{pmatrix} 0 \\ -1 \\ -1 \end{pmatrix},$ $\mathbf{b} = \begin{pmatrix} -4 \\ 4 \\ 4 \end{pmatrix},$ and $\mathbf{c} = \begin{pmatrix} 4 \\ 5 \\ 1 \end{pmatrix}.$  Then the normal vector of the plane is orthogonal to both
\[\mathbf{b} - \mathbf{a} = \begin{pmatrix} -4 \\ 5 \\ 5 \end{pmatrix}\]and
\[\mathbf{c} - \mathbf{a} = \begin{pmatrix} 4 \\ 6 \\ 2 \end{pmatrix}.\]So to compute the normal vector, we take the cross product of these vectors:
\[\begin{pmatrix} -4 \\ 5 \\ 5 \end{pmatrix} \times \begin{pmatrix} 4 \\ 6 \\ 2 \end{pmatrix} = \begin{pmatrix} -20 \\ 28 \\ -44 \end{pmatrix}.\]We can scale this vector, and take $\begin{pmatrix} 5 \\ -7 \\ 11 \end{pmatrix}$ as the normal vector.  Then the equation of the plane is of the form
\[5x - 7y + 11z + D = 0.\]Substituting the coordinates of any of the points, we find that the equation of the plane is $\boxed{5x - 7y + 11z + 4 = 0}.$


Explanation:
Okay, so I need to figure out how to write the expression √2 + 1/√2 + √3 + 1/√3 in the form (a√2 + b√3)/c, where a, b, and c are positive integers, and c is as small as possible. Then, I have to find a + b + c. Alright, let me start by breaking down the problem step by step.

First, let me look at each term individually. The expression has four terms: √2, 1/√2, √3, and 1/√3. Maybe I can combine the terms with √2 and the terms with √3 separately. Let me try that.

Starting with the √2 terms: √2 and 1/√2. If I can combine these two, maybe I can simplify them. Similarly for the √3 terms: √3 and 1/√3. Let me handle them one by one.

For the √2 terms: √2 + 1/√2. To combine these, I need a common denominator. The denominators here are 1 and √2. So, if I multiply the numerator and denominator of √2 by √2, I can write it as (√2 * √2)/√2, which is 2/√2. Wait, but that seems like it's going back to the original term. Maybe I should rationalize the denominator for the 1/√2 term?

Yes, that's probably a better approach. Let me recall that rationalizing denominators involves multiplying numerator and denominator by the same radical to eliminate the radical in the denominator. So, for 1/√2, multiplying numerator and denominator by √2 gives (√2)/2. Therefore, √2 + 1/√2 becomes √2 + √2/2. Now, combining these terms, since they have the same radical, I can add the coefficients. √2 is the same as 2√2/2, so adding 2√2/2 + √2/2 gives (2√2 + √2)/2 = 3√2/2. Okay, so the √2 terms combine to 3√2/2.

Similarly, let me do the same for the √3 terms: √3 + 1/√3. Again, rationalizing the denominator for 1/√3. Multiply numerator and denominator by √3: √3/3. Therefore, √3 + √3/3. Again, combining these terms. √3 is 3√3/3, so adding 3√3/3 + √3/3 gives (3√3 + √3)/3 = 4√3/3. So the √3 terms combine to 4√3/3.

Therefore, the entire original expression simplifies to 3√2/2 + 4√3/3. Now, the problem wants this in the form (a√2 + b√3)/c. So, I need to combine these two fractions into a single fraction. To do that, I need a common denominator. The denominators here are 2 and 3, so the least common denominator is 6.

Let me convert each term to have denominator 6. For 3√2/2, multiply numerator and denominator by 3: (3√2 * 3)/(2 * 3) = 9√2/6. For 4√3/3, multiply numerator and denominator by 2: (4√3 * 2)/(3 * 2) = 8√3/6. Therefore, adding these two fractions together: 9√2/6 + 8√3/6 = (9√2 + 8√3)/6.

So, according to this, the expression is (9√2 + 8√3)/6. Therefore, a is 9, b is 8, and c is 6. The question says that c should be as small as possible. Let me check if this fraction can be simplified further. The numerator is 9√2 + 8√3, and the denominator is 6. Since √2 and √3 are irrational and cannot be combined, the numerator cannot be simplified further. However, we need to check if there is a common factor in the coefficients of the numerator and the denominator.

Looking at the coefficients: 9, 8, and 6. The greatest common divisor (GCD) of 9, 8, and 6. Let me compute GCD step by step. First, GCD of 9 and 8 is 1, since they are consecutive integers. Then, GCD of 1 and 6 is also 1. Therefore, there is no common factor greater than 1 for all three numbers. Therefore, the fraction (9√2 + 8√3)/6 is already in its simplest form with the smallest possible c, which is 6.

Therefore, a = 9, b = 8, c = 6. Therefore, a + b + c = 9 + 8 + 6 = 23. Wait, but let me double-check my steps to make sure I didn't make a mistake.

First, combining √2 + 1/√2. Rationalizing 1/√2 gives √2/2. Then √2 + √2/2. Converting √2 to 2√2/2, adding to √2/2 gives 3√2/2. That seems correct.

Similarly for √3 + 1/√3. Rationalizing gives √3/3. Then √3 is 3√3/3, adding to √3/3 gives 4√3/3. Correct.

Then combining 3√2/2 + 4√3/3. Common denominator 6. Multiply first term by 3/3: 9√2/6. Multiply second term by 2/2: 8√3/6. Adding gives (9√2 + 8√3)/6. Correct.

Check if 9, 8, 6 have a common factor. GCD of 9 and 6 is 3, but since 8 and 3 are coprime, the overall GCD is 1. Therefore, can't reduce further. Therefore, c is 6, which is minimal. Therefore, a + b + c = 9 + 8 + 6 = 23. So the answer is 23?

Wait, but let me check again. Maybe there's another way to combine the terms that results in a smaller denominator? Let me think. Suppose I didn't separate the terms first. Let me try another approach to verify.

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me combine all terms over a common denominator. But that might be more complicated. Alternatively, maybe I can combine them differently?

Alternatively, perhaps I can combine √2 + 1/√2 as I did before, and same with √3 + 1/√3, which I did. So that seems correct. Alternatively, maybe there's a mistake in the calculation?

Wait, let me check the arithmetic again. For √2 + 1/√2:

√2 is approximately 1.414, 1/√2 is approximately 0.707. Adding them gives approx 2.121. Let me compute 3√2 / 2: 3 * 1.414 / 2 ≈ 4.242 / 2 ≈ 2.121. Correct. Similarly for √3 + 1/√3: √3 ≈ 1.732, 1/√3 ≈ 0.577. Adding gives approx 2.309. 4√3 / 3 ≈ 4 * 1.732 / 3 ≈ 6.928 / 3 ≈ 2.309. Correct. Then adding 2.121 + 2.309 ≈ 4.43. Now, let me compute (9√2 + 8√3)/6. Let me compute numerator:

9√2 ≈ 9 * 1.414 ≈ 12.726

8√3 ≈ 8 * 1.732 ≈ 13.856

Adding them gives approx 12.726 + 13.856 ≈ 26.582. Divided by 6: 26.582 / 6 ≈ 4.43. Which matches the previous total. So that seems correct.

Alternatively, if I tried to combine all terms over a common denominator from the start. Let me try that approach to see if I get the same result.

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me write all terms with denominator √2 and √3. But that might not be straightforward. Alternatively, let me consider combining all terms over a common denominator. Let me think.

Alternatively, maybe I can write each term as fractions:

√2 = √2/1, 1/√2, √3 = √3/1, 1/√3. To combine all terms, the common denominator would be... Well, denominators are 1, √2, 1, √3. So the common denominator would be √2 * √3 = √6? But that might complicate things. Let me try.

But maybe that's more complicated. Let me try. Let me write each term with denominator √6:

For √2: √2 = √2 * √6 / √6 = √12 / √6 = (2√3)/√6. Wait, but that seems messy. Alternatively, maybe not the best approach.

Alternatively, perhaps I can combine terms step by step. Let me try adding √2 + 1/√2 first, which we already did as 3√2/2. Similarly, √3 + 1/√3 is 4√3/3. Then adding those two fractions: 3√2/2 + 4√3/3. Which is the same as before. So, same result. Therefore, seems like the same answer.

Alternatively, maybe there's a different way to combine terms? Let me think. Suppose I didn't rationalize the denominators first. Let me try that approach.

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me combine √2 + 1/√2. Let me write them as fractions:

√2 = (√2 * √2)/√2 = 2/√2. Therefore, √2 + 1/√2 = 2/√2 + 1/√2 = 3/√2. Wait, that's another way to think about it. So 3/√2. Then rationalizing that gives 3√2 / 2. Which is the same as before. Similarly for √3 + 1/√3: √3 = 3/√3, so adding 3/√3 + 1/√3 = 4/√3, which rationalizes to 4√3 / 3. So same result. Therefore, same conclusion.

Therefore, combining them gives 3√2/2 + 4√3/3 = (9√2 + 8√3)/6. So same as before. Therefore, seems like the answer is 9 + 8 + 6 = 23.

But let me check again if there's a way to have a smaller denominator. Suppose that there is some common factor that I missed. For example, if the numerator had a common factor with the denominator. Let me check.

The numerator is 9√2 + 8√3. The denominator is 6. Suppose that there is a common factor between 9 and 6, which is 3, but 8 and 3 are coprime. Therefore, if I factor out 3 from the numerator, but since the numerator has 9√2 and 8√3, factoring 3 would give 3*(3√2 + (8/3)√3). But that introduces a fraction, which is not allowed because a, b, c must be integers. Therefore, we can't factor out 3. Similarly, trying to factor out 2: 9 is odd, 8 is even, so no. Therefore, there is no common factor. Therefore, the fraction is reduced completely. Therefore, c = 6 is indeed minimal.

Alternatively, maybe there's another way to combine the original terms? Let me think. Suppose I didn't separate the √2 and √3 terms. Let me try combining all four terms together. Let me write them all over a common denominator. Let me try that.

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me consider combining all terms. Let me think of them as four separate terms. To combine them, I need a common denominator. The denominators are 1, √2, 1, √3. So the common denominator would be √2 * √3 = √6. Let me try converting each term to have denominator √6.

First term: √2. To write this with denominator √6, multiply numerator and denominator by √6: √2 * √6 / √6 = √12 / √6 = (2√3)/√6. Wait, but that seems messy. Let me check:

Wait, √2 = √2 * (√6 / √6) = √(2*6)/√6 = √12 / √6 = (2√3)/√6. Similarly, 1/√2 = 1/√2 * √6/√6 = √6 / (√2 * √6) = √6 / √12 = √6 / (2√3) = (√6 / √3) / 2 = √(6/3)/2 = √2 / 2. Wait, that seems like going back to the previous step. Maybe this approach is not helpful.

Alternatively, maybe I should convert each term to have denominator √6:

First term: √2 = √2 * (√6 / √6) = √(12)/√6 = 2√3 / √6.

Second term: 1/√2 = 1/√2 * √6 / √6 = √6 / √12 = √6 / (2√3) = (√6 / √3) / 2 = √2 / 2. Wait, again, same as before. So maybe this isn't helpful. Let me try to write all terms over √6:

First term: √2 = 2√3 / √6.

Second term: 1/√2 = √2 / 2. Wait, but how to write that over √6? Let me think. If I want denominator √6, then √2 / 2 = (√2 * √6) / (2 * √6) = √12 / (2√6) = (2√3) / (2√6) = √3 / √6. So that's another way. Therefore, second term is √3 / √6.

Third term: √3. Similarly, √3 = √3 * √6 / √6 = √18 / √6 = (3√2)/√6.

Fourth term: 1/√3 = 1/√3 * √6 / √6 = √6 / √18 = √6 / (3√2) = (√6 / √2) / 3 = √3 / 3. Wait, but again, how to write that over √6? Let me do the same as before. 1/√3 = √3 / 3. To write over √6, multiply numerator and denominator by √6: (√3 * √6) / (3 * √6) = √18 / (3√6) = (3√2) / (3√6) = √2 / √6.

Therefore, fourth term is √2 / √6.

So putting all terms together:

First term: 2√3 / √6

Second term: √3 / √6

Third term: 3√2 / √6

Fourth term: √2 / √6

Adding them all together: (2√3 + √3 + 3√2 + √2)/√6 = (3√3 + 4√2)/√6. Then, combining numerator and denominator:

(4√2 + 3√3)/√6. Now, to rationalize the denominator here, multiply numerator and denominator by √6:

[(4√2 + 3√3) * √6] / (√6 * √6) = [(4√2 * √6 + 3√3 * √6)] / 6.

Let me compute the numerator:

First term: 4√2 * √6 = 4√(12) = 4 * 2√3 = 8√3.

Second term: 3√3 * √6 = 3√(18) = 3 * 3√2 = 9√2.

Therefore, numerator is 8√3 + 9√2, denominator is 6. Therefore, the expression becomes (9√2 + 8√3)/6, which is the same as before. Therefore, same result. Therefore, regardless of the method, I end up with (9√2 + 8√3)/6. Therefore, confirming that the answer is 9 + 8 + 6 = 23.

But just to be thorough, let me check if there's a different combination where c is smaller. For example, suppose that there is a common factor between the numerator terms and denominator. But since the numerator is 9√2 + 8√3, and denominator is 6. If there was a common factor, say, 2, but 9 is odd, so 9√2 can't be divided by 2. Similarly, 8 is divisible by 2, but 9 isn't. Therefore, we can't factor out 2. Similarly, 3 divides 9 and 6, but not 8. Therefore, if we tried to factor out 3, we would have 3*(3√2 + (8/3)√3)/6, which is 3*(something)/6 = (something)/2, but then the numerator would have fractions, which is not allowed. Therefore, no, we can't do that. Therefore, 6 is indeed the minimal denominator.

Alternatively, maybe there's a different way to combine the original terms? Let me think. Suppose I didn't rationalize the denominators first. Let me try adding all terms as they are:

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me combine √2 + 1/√2 and √3 + 1/√3 as before, but maybe there's another way? For example, if I consider √2 + 1/√2 = (√2 * √2 + 1)/√2 = (2 + 1)/√2 = 3/√2. Wait, that's another way to think about it. Similarly, √3 + 1/√3 = (√3 * √3 + 1)/√3 = (3 + 1)/√3 = 4/√3. Therefore, the entire expression is 3/√2 + 4/√3. Then, combining these two terms. To add 3/√2 + 4/√3, we need a common denominator. The common denominator would be √2 * √3 = √6. Therefore, converting each term:

3/√2 = 3√3 / √6

4/√3 = 4√2 / √6

Adding them together: (3√3 + 4√2)/√6. Then, rationalizing the denominator as before gives (3√3 + 4√2) * √6 / 6 = same as before, which gives (9√2 + 8√3)/6. So same result. Therefore, regardless of the method, same answer.

Therefore, I think I can be confident that the expression simplifies to (9√2 + 8√3)/6, so a = 9, b = 8, c = 6, and a + b + c = 23.

But just to make sure, let me check if there's a different form where c is smaller. For example, suppose that someone might think that combining terms differently could lead to a smaller denominator. Let me think. For example, if I consider the entire original expression:

√2 + 1/√2 + √3 + 1/√3. Suppose I write all terms with denominator 2 and 3. Wait, but that's what I did before. Alternatively, maybe there's a way to combine them in a different way?

Alternatively, suppose I consider the entire expression as:

(√2 + 1/√2) + (√3 + 1/√3) = 3√2/2 + 4√3/3. If I wanted to write this as a single fraction, the denominator is 6, so that's the same as before. Therefore, no way around that. Therefore, c = 6 is indeed minimal.

Alternatively, if there was a way to have a different combination where the denominator is smaller. For example, if the numerator had some cancellation. But since the numerator is 9√2 + 8√3, which are separate radicals, they can't be combined further. Therefore, the denominator can't be reduced. Therefore, 6 is the minimal possible denominator.

Therefore, I think my answer is correct. Therefore, the answer is 23.

**Final Answer**
\boxed{23}

To simplify the expression
$$
\sqrt{2} + \frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}
$$
into the form
$$
\frac{a\sqrt{2} + b\sqrt{3}}{c}
$$
with $ a, b, c $ as positive integers and $ c $ as small as possible, we proceed step by step.

---

### Step 1: Simplify each pair of terms

We group the terms with the same radical:

- **For $ \sqrt{2} $:**
  $$
  \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}
  $$
  Rationalizing the denominator:
  $$
  \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}
  $$

- **For $ \sqrt{3} $:**
  $$
  \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}
  $$
  Rationalizing the denominator:
  $$
  \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}
  $$

---

### Step 2: Combine the simplified terms

Now we have:
$$
\frac{3\sqrt{2}}{2} + \frac{4\sqrt{3}}{3}
$$

To add these, we find a common denominator. The least common denominator of 2 and 3 is 6.

- Convert $ \frac{3\sqrt{2}}{2} $ to denominator 6:
  $$
  \frac{3\sqrt{2}}{2} = \frac{9\sqrt{2}}{6}
  $$

- Convert $ \frac{4\sqrt{3}}{3} $ to denominator 6:
  $$
  \frac{4\sqrt{3}}{3} = \frac{8\sqrt{3}}{6}
  $$

Now add the two:
$$
\frac{9\sqrt{2}}{6} + \frac{8\sqrt{3}}{6} = \frac{9\sqrt{2} + 8\sqrt{3}}{6}
$$

---

### Step 3: Identify $ a, b, c $

From the expression:
$$
\frac{9\sqrt{2} + 8\sqrt{3}}{6}
$$
we identify:
- $ a = 9 $
- $ b = 8 $
- $ c = 6 $

We check if $ c $ can be smaller. Since the numerator contains two distinct radicals, $ \sqrt{2} $ and $ \sqrt{3} $, they cannot be combined further. The coefficients 9, 8, and 6 have a greatest common divisor of 1, so the fraction is in its simplest form.

---

### Final Step: Compute $ a + b + c $

$$
a + b + c = 9 + 8 + 6 = \boxed{23}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how to write the expression √2 + 1/√2 + √3 + 1/√3 in the form (a√2 + b√3)/c, where a, b, and c are positive integers, and c is as small as possible. Then, I have to find a + b + c. Alright, let me start by breaking down the problem step by step.

First, let me look at each term individually. The expression has four terms: √2, 1/√2, √3, and 1/√3. Maybe I can combine the terms with √2 and the terms with √3 separately. Let me try that.

Starting with the √2 terms: √2 and 1/√2. If I can combine these two, maybe I can simplify them. Similarly for the √3 terms: √3 and 1/√3. Let me handle them one by one.

For the √2 terms: √2 + 1/√2. To combine these, I need a common denominator. The denominators here are 1 and √2. So, if I multiply the numerator and denominator of √2 by √2, I can write it as (√2 * √2)/√2, which is 2/√2. Wait, but that seems like it's going back to the original term. Maybe I should rationalize the denominator for the 1/√2 term?

Yes, that's probably a better approach. Let me recall that rationalizing denominators involves multiplying numerator and denominator by the same radical to eliminate the radical in the denominator. So, for 1/√2, multiplying numerator and denominator by √2 gives (√2)/2. Therefore, √2 + 1/√2 becomes √2 + √2/2. Now, combining these terms, since they have the same radical, I can add the coefficients. √2 is the same as 2√2/2, so adding 2√2/2 + √2/2 gives (2√2 + √2)/2 = 3√2/2. Okay, so the √2 terms combine to 3√2/2.

Similarly, let me do the same for the √3 terms: √3 + 1/√3. Again, rationalizing the denominator for 1/√3. Multiply numerator and denominator by √3: √3/3. Therefore, √3 + √3/3. Again, combining these terms. √3 is 3√3/3, so adding 3√3/3 + √3/3 gives (3√3 + √3)/3 = 4√3/3. So the √3 terms combine to 4√3/3.

Therefore, the entire original expression simplifies to 3√2/2 + 4√3/3. Now, the problem wants this in the form (a√2 + b√3)/c. So, I need to combine these two fractions into a single fraction. To do that, I need a common denominator. The denominators here are 2 and 3, so the least common denominator is 6.

Let me convert each term to have denominator 6. For 3√2/2, multiply numerator and denominator by 3: (3√2 * 3)/(2 * 3) = 9√2/6. For 4√3/3, multiply numerator and denominator by 2: (4√3 * 2)/(3 * 2) = 8√3/6. Therefore, adding these two fractions together: 9√2/6 + 8√3/6 = (9√2 + 8√3)/6.

So, according to this, the expression is (9√2 + 8√3)/6. Therefore, a is 9, b is 8, and c is 6. The question says that c should be as small as possible. Let me check if this fraction can be simplified further. The numerator is 9√2 + 8√3, and the denominator is 6. Since √2 and √3 are irrational and cannot be combined, the numerator cannot be simplified further. However, we need to check if there is a common factor in the coefficients of the numerator and the denominator.

Looking at the coefficients: 9, 8, and 6. The greatest common divisor (GCD) of 9, 8, and 6. Let me compute GCD step by step. First, GCD of 9 and 8 is 1, since they are consecutive integers. Then, GCD of 1 and 6 is also 1. Therefore, there is no common factor greater than 1 for all three numbers. Therefore, the fraction (9√2 + 8√3)/6 is already in its simplest form with the smallest possible c, which is 6.

Therefore, a = 9, b = 8, c = 6. Therefore, a + b + c = 9 + 8 + 6 = 23. Wait, but let me double-check my steps to make sure I didn't make a mistake.

First, combining √2 + 1/√2. Rationalizing 1/√2 gives √2/2. Then √2 + √2/2. Converting √2 to 2√2/2, adding to √2/2 gives 3√2/2. That seems correct.

Similarly for √3 + 1/√3. Rationalizing gives √3/3. Then √3 is 3√3/3, adding to √3/3 gives 4√3/3. Correct.

Then combining 3√2/2 + 4√3/3. Common denominator 6. Multiply first term by 3/3: 9√2/6. Multiply second term by 2/2: 8√3/6. Adding gives (9√2 + 8√3)/6. Correct.

Check if 9, 8, 6 have a common factor. GCD of 9 and 6 is 3, but since 8 and 3 are coprime, the overall GCD is 1. Therefore, can't reduce further. Therefore, c is 6, which is minimal. Therefore, a + b + c = 9 + 8 + 6 = 23. So the answer is 23?

Wait, but let me check again. Maybe there's another way to combine the terms that results in a smaller denominator? Let me think. Suppose I didn't separate the terms first. Let me try another approach to verify.

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me combine all terms over a common denominator. But that might be more complicated. Alternatively, maybe I can combine them differently?

Alternatively, perhaps I can combine √2 + 1/√2 as I did before, and same with √3 + 1/√3, which I did. So that seems correct. Alternatively, maybe there's a mistake in the calculation?

Wait, let me check the arithmetic again. For √2 + 1/√2:

√2 is approximately 1.414, 1/√2 is approximately 0.707. Adding them gives approx 2.121. Let me compute 3√2 / 2: 3 * 1.414 / 2 ≈ 4.242 / 2 ≈ 2.121. Correct. Similarly for √3 + 1/√3: √3 ≈ 1.732, 1/√3 ≈ 0.577. Adding gives approx 2.309. 4√3 / 3 ≈ 4 * 1.732 / 3 ≈ 6.928 / 3 ≈ 2.309. Correct. Then adding 2.121 + 2.309 ≈ 4.43. Now, let me compute (9√2 + 8√3)/6. Let me compute numerator:

9√2 ≈ 9 * 1.414 ≈ 12.726

8√3 ≈ 8 * 1.732 ≈ 13.856

Adding them gives approx 12.726 + 13.856 ≈ 26.582. Divided by 6: 26.582 / 6 ≈ 4.43. Which matches the previous total. So that seems correct.

Alternatively, if I tried to combine all terms over a common denominator from the start. Let me try that approach to see if I get the same result.

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me write all terms with denominator √2 and √3. But that might not be straightforward. Alternatively, let me consider combining all terms over a common denominator. Let me think.

Alternatively, maybe I can write each term as fractions:

√2 = √2/1, 1/√2, √3 = √3/1, 1/√3. To combine all terms, the common denominator would be... Well, denominators are 1, √2, 1, √3. So the common denominator would be √2 * √3 = √6? But that might complicate things. Let me try.

But maybe that's more complicated. Let me try. Let me write each term with denominator √6:

For √2: √2 = √2 * √6 / √6 = √12 / √6 = (2√3)/√6. Wait, but that seems messy. Alternatively, maybe not the best approach.

Alternatively, perhaps I can combine terms step by step. Let me try adding √2 + 1/√2 first, which we already did as 3√2/2. Similarly, √3 + 1/√3 is 4√3/3. Then adding those two fractions: 3√2/2 + 4√3/3. Which is the same as before. So, same result. Therefore, seems like the same answer.

Alternatively, maybe there's a different way to combine terms? Let me think. Suppose I didn't rationalize the denominators first. Let me try that approach.

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me combine √2 + 1/√2. Let me write them as fractions:

√2 = (√2 * √2)/√2 = 2/√2. Therefore, √2 + 1/√2 = 2/√2 + 1/√2 = 3/√2. Wait, that's another way to think about it. So 3/√2. Then rationalizing that gives 3√2 / 2. Which is the same as before. Similarly for √3 + 1/√3: √3 = 3/√3, so adding 3/√3 + 1/√3 = 4/√3, which rationalizes to 4√3 / 3. So same result. Therefore, same conclusion.

Therefore, combining them gives 3√2/2 + 4√3/3 = (9√2 + 8√3)/6. So same as before. Therefore, seems like the answer is 9 + 8 + 6 = 23.

But let me check again if there's a way to have a smaller denominator. Suppose that there is some common factor that I missed. For example, if the numerator had a common factor with the denominator. Let me check.

The numerator is 9√2 + 8√3. The denominator is 6. Suppose that there is a common factor between 9 and 6, which is 3, but 8 and 3 are coprime. Therefore, if I factor out 3 from the numerator, but since the numerator has 9√2 and 8√3, factoring 3 would give 3*(3√2 + (8/3)√3). But that introduces a fraction, which is not allowed because a, b, c must be integers. Therefore, we can't factor out 3. Similarly, trying to factor out 2: 9 is odd, 8 is even, so no. Therefore, there is no common factor. Therefore, the fraction is reduced completely. Therefore, c = 6 is indeed minimal.

Alternatively, maybe there's another way to combine the original terms? Let me think. Suppose I didn't separate the √2 and √3 terms. Let me try combining all four terms together. Let me write them all over a common denominator. Let me try that.

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me consider combining all terms. Let me think of them as four separate terms. To combine them, I need a common denominator. The denominators are 1, √2, 1, √3. So the common denominator would be √2 * √3 = √6. Let me try converting each term to have denominator √6.

First term: √2. To write this with denominator √6, multiply numerator and denominator by √6: √2 * √6 / √6 = √12 / √6 = (2√3)/√6. Wait, but that seems messy. Let me check:

Wait, √2 = √2 * (√6 / √6) = √(2*6)/√6 = √12 / √6 = (2√3)/√6. Similarly, 1/√2 = 1/√2 * √6/√6 = √6 / (√2 * √6) = √6 / √12 = √6 / (2√3) = (√6 / √3) / 2 = √(6/3)/2 = √2 / 2. Wait, that seems like going back to the previous step. Maybe this approach is not helpful.

Alternatively, maybe I should convert each term to have denominator √6:

First term: √2 = √2 * (√6 / √6) = √(12)/√6 = 2√3 / √6.

Second term: 1/√2 = 1/√2 * √6 / √6 = √6 / √12 = √6 / (2√3) = (√6 / √3) / 2 = √2 / 2. Wait, again, same as before. So maybe this isn't helpful. Let me try to write all terms over √6:

First term: √2 = 2√3 / √6.

Second term: 1/√2 = √2 / 2. Wait, but how to write that over √6? Let me think. If I want denominator √6, then √2 / 2 = (√2 * √6) / (2 * √6) = √12 / (2√6) = (2√3) / (2√6) = √3 / √6. So that's another way. Therefore, second term is √3 / √6.

Third term: √3. Similarly, √3 = √3 * √6 / √6 = √18 / √6 = (3√2)/√6.

Fourth term: 1/√3 = 1/√3 * √6 / √6 = √6 / √18 = √6 / (3√2) = (√6 / √2) / 3 = √3 / 3. Wait, but again, how to write that over √6? Let me do the same as before. 1/√3 = √3 / 3. To write over √6, multiply numerator and denominator by √6: (√3 * √6) / (3 * √6) = √18 / (3√6) = (3√2) / (3√6) = √2 / √6.

Therefore, fourth term is √2 / √6.

So putting all terms together:

First term: 2√3 / √6

Second term: √3 / √6

Third term: 3√2 / √6

Fourth term: √2 / √6

Adding them all together: (2√3 + √3 + 3√2 + √2)/√6 = (3√3 + 4√2)/√6. Then, combining numerator and denominator:

(4√2 + 3√3)/√6. Now, to rationalize the denominator here, multiply numerator and denominator by √6:

[(4√2 + 3√3) * √6] / (√6 * √6) = [(4√2 * √6 + 3√3 * √6)] / 6.

Let me compute the numerator:

First term: 4√2 * √6 = 4√(12) = 4 * 2√3 = 8√3.

Second term: 3√3 * √6 = 3√(18) = 3 * 3√2 = 9√2.

Therefore, numerator is 8√3 + 9√2, denominator is 6. Therefore, the expression becomes (9√2 + 8√3)/6, which is the same as before. Therefore, same result. Therefore, regardless of the method, I end up with (9√2 + 8√3)/6. Therefore, confirming that the answer is 9 + 8 + 6 = 23.

But just to be thorough, let me check if there's a different combination where c is smaller. For example, suppose that there is a common factor between the numerator terms and denominator. But since the numerator is 9√2 + 8√3, and denominator is 6. If there was a common factor, say, 2, but 9 is odd, so 9√2 can't be divided by 2. Similarly, 8 is divisible by 2, but 9 isn't. Therefore, we can't factor out 2. Similarly, 3 divides 9 and 6, but not 8. Therefore, if we tried to factor out 3, we would have 3*(3√2 + (8/3)√3)/6, which is 3*(something)/6 = (something)/2, but then the numerator would have fractions, which is not allowed. Therefore, no, we can't do that. Therefore, 6 is indeed the minimal denominator.

Alternatively, maybe there's a different way to combine the original terms? Let me think. Suppose I didn't rationalize the denominators first. Let me try adding all terms as they are:

Original expression: √2 + 1/√2 + √3 + 1/√3. Let me combine √2 + 1/√2 and √3 + 1/√3 as before, but maybe there's another way? For example, if I consider √2 + 1/√2 = (√2 * √2 + 1)/√2 = (2 + 1)/√2 = 3/√2. Wait, that's another way to think about it. Similarly, √3 + 1/√3 = (√3 * √3 + 1)/√3 = (3 + 1)/√3 = 4/√3. Therefore, the entire expression is 3/√2 + 4/√3. Then, combining these two terms. To add 3/√2 + 4/√3, we need a common denominator. The common denominator would be √2 * √3 = √6. Therefore, converting each term:

3/√2 = 3√3 / √6

4/√3 = 4√2 / √6

Adding them together: (3√3 + 4√2)/√6. Then, rationalizing the denominator as before gives (3√3 + 4√2) * √6 / 6 = same as before, which gives (9√2 + 8√3)/6. So same result. Therefore, regardless of the method, same answer.

Therefore, I think I can be confident that the expression simplifies to (9√2 + 8√3)/6, so a = 9, b = 8, c = 6, and a + b + c = 23.

But just to make sure, let me check if there's a different form where c is smaller. For example, suppose that someone might think that combining terms differently could lead to a smaller denominator. Let me think. For example, if I consider the entire original expression:

√2 + 1/√2 + √3 + 1/√3. Suppose I write all terms with denominator 2 and 3. Wait, but that's what I did before. Alternatively, maybe there's a way to combine them in a different way?

Alternatively, suppose I consider the entire expression as:

(√2 + 1/√2) + (√3 + 1/√3) = 3√2/2 + 4√3/3. If I wanted to write this as a single fraction, the denominator is 6, so that's the same as before. Therefore, no way around that. Therefore, c = 6 is indeed minimal.

Alternatively, if there was a way to have a different combination where the denominator is smaller. For example, if the numerator had some cancellation. But since the numerator is 9√2 + 8√3, which are separate radicals, they can't be combined further. Therefore, the denominator can't be reduced. Therefore, 6 is the minimal possible denominator.

Therefore, I think my answer is correct. Therefore, the answer is 23.

**Final Answer**
\boxed{23}

To simplify the expression
$$
\sqrt{2} + \frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}
$$
into the form
$$
\frac{a\sqrt{2} + b\sqrt{3}}{c}
$$
with $ a, b, c $ as positive integers and $ c $ as small as possible, we proceed step by step.

---

### Step 1: Simplify each pair of terms

We group the terms with the same radical:

- **For $ \sqrt{2} $:**
  $$
  \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}
  $$
  Rationalizing the denominator:
  $$
  \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}
  $$

- **For $ \sqrt{3} $:**
  $$
  \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}
  $$
  Rationalizing the denominator:
  $$
  \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}
  $$

---

### Step 2: Combine the simplified terms

Now we have:
$$
\frac{3\sqrt{2}}{2} + \frac{4\sqrt{3}}{3}
$$

To add these, we find a common denominator. The least common denominator of 2 and 3 is 6.

- Convert $ \frac{3\sqrt{2}}{2} $ to denominator 6:
  $$
  \frac{3\sqrt{2}}{2} = \frac{9\sqrt{2}}{6}
  $$

- Convert $ \frac{4\sqrt{3}}{3} $ to denominator 6:
  $$
  \frac{4\sqrt{3}}{3} = \frac{8\sqrt{3}}{6}
  $$

Now add the two:
$$
\frac{9\sqrt{2}}{6} + \frac{8\sqrt{3}}{6} = \frac{9\sqrt{2} + 8\sqrt{3}}{6}
$$

---

### Step 3: Identify $ a, b, c $

From the expression:
$$
\frac{9\sqrt{2} + 8\sqrt{3}}{6}
$$
we identify:
- $ a = 9 $
- $ b = 8 $
- $ c = 6 $

We check if $ c $ can be smaller. Since the numerator contains two distinct radicals, $ \sqrt{2} $ and $ \sqrt{3} $, they cannot be combined further. The coefficients 9, 8, and 6 have a greatest common divisor of 1, so the fraction is in its simplest form.

---

### Final Step: Compute $ a + b + c $

$$
a + b + c = 9 + 8 + 6 = \boxed{23}
$$

Solution: The common denominator desired is $\sqrt{2}\cdot\sqrt{3} = \sqrt{6}$. So, this expression becomes $\frac{\sqrt{2}\cdot(\sqrt{2}\cdot\sqrt{3})+1\cdot\sqrt{3}+\sqrt{3}\cdot(\sqrt{2}\cdot\sqrt{3})+1\cdot\sqrt{2}}{\sqrt{6}}$. Simplifying this gives $\frac{2\sqrt{3}+\sqrt{3}+3\sqrt{2}+\sqrt{2}}{\sqrt{6}} = \frac{4\sqrt{2}+3\sqrt{3}}{\sqrt{6}}$. To rationalize, multiply numerator and denominator by $\sqrt{6}$ to get $\frac{4\sqrt{2}\sqrt{6}+3\sqrt{3}\sqrt{6}}{6}$. Simplifying yields ${\frac{9\sqrt{2}+8\sqrt{3}}{6}}$, so the desired sum is $9+8+6=\boxed{23}$.