Explanation:
Okay, so I need to figure out after how many seconds both gears will have their marks pointing due north again. Let me start by understanding the problem.

First, there are two gears. One turns 33 and 1/3 times per minute, and the other turns 45 times per minute. Initially, both have a mark pointing due north. I need to find the time when both marks will point north again simultaneously. The answer should be in seconds.

Hmm, gears... when gears are meshed together, their rotations are related. But wait, the problem doesn't mention them being connected. Wait, but maybe they are separate gears? Or maybe they are connected? Wait, the problem says "a mark on each gear is pointing due north." So maybe they are separate gears, not connected? But then, how does their rotation affect each other? Wait, maybe they are connected? Wait, the problem doesn't specify. Hmm. Wait, but maybe it's not necessary? Maybe it's just about the rotation of each gear independently?

Wait, but if they are separate gears, then their marks pointing north would depend on their own rotation. So, for each gear, the mark will point north every time it completes a full rotation. So, for the first gear, which turns 33 1/3 times per minute, how long does it take for the mark to point north again? Similarly for the second gear. Then, the question is when both will have their marks pointing north again, which would be the least common multiple (LCM) of their individual periods.

But let me think again. If they are separate gears, then their marks will point north again after their individual periods. So, the first gear's period is 1 / (33 1/3) minutes per rotation, and the second gear's period is 1 / 45 minutes per rotation. Then, the time when both marks point north again is the LCM of these two periods. But since LCM is usually for integers, maybe I need to convert them into fractions?

Alternatively, maybe think in terms of how many rotations each gear makes in a certain time, and find when both have completed an integer number of rotations. So, the time when both marks point north again is the smallest time t such that t multiplied by the number of rotations per minute for each gear is an integer. So, for the first gear: t * 33 1/3 must be integer, and for the second gear: t * 45 must be integer. Then, find the smallest t (in minutes) that satisfies both, then convert to seconds.

Yes, that seems like a better approach. Let me formalize this.

Let me denote:

For gear 1: rotation rate is 33 1/3 rotations per minute. Let me convert that to an improper fraction. 33 1/3 is equal to 100/3 rotations per minute. So, 100/3 rpm.

For gear 2: 45 rotations per minute.

So, if I let t be the time in minutes, then the number of rotations for gear 1 is (100/3) * t, and for gear 2 is 45 * t. We need both of these numbers to be integers. So, (100/3)*t must be integer, and 45*t must be integer.

So, let me write equations:

(100/3)*t ∈ ℤ

and

45*t ∈ ℤ

So, t must be such that t is a multiple of 3/100 minutes to make (100/3)*t integer. Wait, let me think. Let me solve for t.

For the first equation: (100/3)*t is integer. Let me denote that as:

Let me write t = k / (100/3) where k is integer? Wait, no. Let me think again. If (100/3)*t is integer, then t must be a multiple of 3/100. Because if you have (100/3)*t = integer, then t = (integer) * 3/100. Similarly, for the second equation, 45*t is integer, so t must be a multiple of 1/45.

Therefore, t must be a common multiple of 3/100 and 1/45. Therefore, the least common multiple of these two fractions. How do I compute LCM of fractions?

I remember that LCM of fractions is LCM(numerators)/GCD(denominators). Wait, or is it LCM(numerators)/GCD(denominators)? Wait, actually, the formula for LCM of fractions is LCM of numerators divided by GCD of denominators. Wait, let me check.

Alternatively, another way: Let me convert the fractions to have the same denominator. Let me think. Let me recall that LCM(a/b, c/d) = LCM(a, c) / GCD(b, d). Wait, not sure. Maybe better to think in terms of converting to decimals?

Alternatively, think of t as a multiple of both 3/100 and 1/45. So, t must be a common multiple of 3/100 and 1/45. So, to find the least such t, we can find LCM of 3/100 and 1/45. How?

Alternatively, think of t as a number such that t = (3/100)*k = (1/45)*m, where k and m are integers. So, solving for t, we can write:

(3/100)*k = (1/45)*m => m = (3/100)*k * 45 = (135/100)*k = (27/20)*k

Since m must be integer, then (27/20)*k must be integer. Therefore, k must be a multiple of 20. Let me denote k = 20*n, where n is integer. Then, m = (27/20)*(20*n) = 27*n. Therefore, t = (3/100)*k = (3/100)*(20*n) = (60/100)*n = (3/5)*n. Therefore, the minimal positive t is when n=1, so t= 3/5 minutes. Wait, but let me check if this is correct.

Wait, so according to this, t= 3/5 minutes. Let me check if this satisfies both conditions.

First, for gear 1: (100/3)*t = (100/3)*(3/5) = 100/5 = 20, which is integer. Good.

For gear 2: 45*t = 45*(3/5) = 27, which is integer. Good. So, t= 3/5 minutes is the minimal time? Wait, but is there a smaller t?

Wait, let me think again. Because if I take LCM of 3/100 and 1/45, maybe there is a smaller t. Let me try another approach.

Let me convert both periods into fractions. The period for gear 1 is 1 / (100/3) = 3/100 minutes per rotation. The period for gear 2 is 1 / 45 minutes per rotation. So, the question is, when will both gears have completed an integer number of rotations? So, the time t must be a multiple of both 3/100 and 1/45. Therefore, t is the LCM of 3/100 and 1/45.

To compute LCM of two fractions, the formula is LCM(numerator1, numerator2) / GCD(denominator1, denominator2). Wait, but I need to confirm.

Alternatively, LCM(a, b) = |a*b| / GCD(a, b). But that is for integers. For fractions, perhaps?

Alternatively, think of LCM of two numbers as the smallest number that is a multiple of both. So, if I have two numbers, say, 3/100 and 1/45, then LCM is the smallest t such that t is a multiple of both. So, t = (3/100)*k = (1/45)*m. Which is the same as before.

Alternatively, convert them to have the same denominator. Let me find a common denominator for 3/100 and 1/45. The denominators are 100 and 45. The LCM of 100 and 45 is... Let me compute LCM(100, 45). Prime factors of 100: 2^2 * 5^2. Prime factors of 45: 3^2 * 5. So LCM is 2^2 * 3^2 * 5^2 = 4 * 9 * 25 = 900. So, common denominator is 900.

So, 3/100 = (3 * 9)/900 = 27/900

1/45 = (20)/900

So, now, LCM of 27/900 and 20/900 is LCM(27, 20)/900. Since denominators are same. LCM(27, 20). 27 is 3^3, 20 is 2^2 * 5. So LCM is 2^2 * 3^3 * 5 = 4 * 27 * 5 = 4 * 135 = 540. Therefore, LCM is 540 / 900 = 3/5 minutes. Which matches the previous result. So, t= 3/5 minutes. Therefore, converting to seconds, 3/5 minutes is 3/5 * 60 seconds = 36 seconds. Therefore, the answer is 36 seconds?

Wait, but let me check again. Let me verify with another approach.

Alternatively, think in terms of how many rotations each gear makes in t seconds. Wait, but maybe better to think in terms of how many rotations per second?

Wait, let me try again. Let me convert the rotation rates to rotations per second, so that I can work in seconds.

First gear: 33 1/3 rotations per minute. 33 1/3 is 100/3 rotations per minute. So, per second, that is (100/3)/60 = 100/(3*60) = 100/180 = 5/9 rotations per second.

Second gear: 45 rotations per minute, which is 45/60 = 3/4 rotations per second.

So, the first gear makes 5/9 rotations per second, the second makes 3/4 rotations per second.

We need to find the smallest t (in seconds) such that both gears have completed an integer number of rotations. So:

For gear 1: (5/9)*t must be integer.

For gear 2: (3/4)*t must be integer.

So, solving for t:

Let me denote t must satisfy:

(5/9)*t ∈ ℤ and (3/4)*t ∈ ℤ.

So, t must be such that t is a multiple of 9/5 and 4/3. Wait, similar to before. So, t must be a common multiple of 9/5 and 4/3. Therefore, LCM of 9/5 and 4/3.

Again, using the same method as before. Let me convert them to fractions with common denominators or use the formula.

Alternatively, find t such that t = (9/5)*k = (4/3)*m, where k and m are integers. Then:

(9/5)*k = (4/3)*m => 27k = 20m => 27k = 20m. Therefore, 27 divides 20m. Since GCD(27,20)=1, 27 divides m. Let m=27n. Then, 27k = 20*27n => k=20n. Therefore, t = (9/5)*k = (9/5)*20n = 36n. Therefore, the minimal t is 36 seconds when n=1. Therefore, the answer is 36 seconds. Which matches the previous result. So, that seems consistent.

Therefore, the answer is 36 seconds. Let me check once more with actual numbers.

First gear: 5/9 rotations per second. After 36 seconds, number of rotations is 5/9 * 36 = 20 rotations. Which is integer. Good.

Second gear: 3/4 rotations per second. After 36 seconds, number of rotations is 3/4 * 36 = 27 rotations. Which is integer. Perfect. So, both gears have completed integer number of rotations, so their marks point north again. And since 36 is the minimal such time, that's the answer.

But let me check if there is a smaller time. Suppose, for example, 18 seconds. Let's check:

First gear: 5/9 * 18 = 10 rotations. Integer. Second gear: 3/4 * 18 = 13.5 rotations. Not integer. So, no. How about 24 seconds?

First gear: 5/9 *24 = 13.333... Not integer. Second gear: 3/4 *24 = 18. So, only second gear is integer. Not both. How about 12 seconds?

First gear: 5/9 *12 = 6.666... Not integer. Second gear: 3/4 *12 = 9. Not both. How about 36 seconds? As above, both integers. What about 18 seconds? As above, only second gear. How about 9 seconds?

First gear: 5/9 *9 = 5. Second gear: 3/4 *9 = 6.75. Not integer. So, no. How about 6 seconds?

First gear: 5/9 *6 = 10/3 ≈ 3.333. Not integer. Second gear: 3/4 *6 = 4.5. Not integer. How about 27 seconds?

First gear: 5/9 *27 = 15. Second gear: 3/4 *27 = 20.25. Not integer. How about 36 seconds? As before, both integers. So, seems like 36 is the minimal. Therefore, answer is 36 seconds.

Alternatively, another approach: find the LCM of the periods. The period of gear 1 is 3/100 minutes, period of gear 2 is 1/45 minutes. So, LCM(3/100, 1/45). As we calculated before, LCM is 3/5 minutes, which is 36 seconds. So, same answer.

Alternatively, think in terms of how many minutes it takes for both gears to align. Since gear 1 makes 100/3 rotations per minute, gear 2 makes 45 rotations per minute. The time when both marks point north again is when both have made an integer number of rotations. So, the number of rotations for gear 1 is (100/3)*t, and for gear 2 is 45*t. So, we need (100/3)*t and 45*t to be integers. Let me write equations:

Let me denote t as minutes. So:

(100/3)*t = integer => t = (3/100)*integer.

Similarly, 45*t = integer => t = (1/45)*integer.

Therefore, t must be a common multiple of 3/100 and 1/45. So, the LCM of these two fractions. As before, LCM(3/100, 1/45) = LCM(3,1)/GCD(100,45) ? Wait, no. Wait, earlier we had LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). Wait, let me check.

Wait, if I have two fractions, say, a/b and c/d, then LCM(a/b, c/d) is LCM(a,c)/GCD(b,d). Wait, but I need to verify.

Alternatively, think of LCM of two fractions as the smallest positive number that is a multiple of both fractions. So, if I have two numbers, say, 3/100 and 1/45, then LCM is the smallest t such that t = 3/100 * k = 1/45 * m, where k and m are integers. Which is the same as before. Which led us to t=3/5 minutes. So, 3/5 minutes is 36 seconds.

Alternatively, think of the problem as finding t such that t is a multiple of both periods. So, the periods are 3/100 minutes and 1/45 minutes. So, LCM(3/100, 1/45). To compute LCM of two numbers, you can use the formula LCM(a,b) = |a*b| / GCD(a,b). But this formula is for integers. For fractions, perhaps?

Wait, if I consider the periods as numbers, then LCM(a,b) = LCM(numerators)/GCD(denominators) ?

Wait, let me check with the previous numbers. For 3/100 and 1/45. If I take LCM of numerators 3 and 1, which is 3. GCD of denominators 100 and 45 is 5. Therefore, LCM would be 3 / 5 = 3/5. Which is exactly what we had. So, that seems to be the formula. Therefore, LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). Therefore, in this case, LCM(3,1)/GCD(100,45) = 3 / 5. Therefore, LCM is 3/5 minutes. Which is 36 seconds. So, that's another way to get the same answer.

Therefore, regardless of the method, the answer is 36 seconds. Therefore, I think that is the correct answer.

But just to make sure, let me think if there's another way someone might approach this problem and maybe get confused.

For example, someone might think that since gear 1 turns 33 1/3 times per minute, which is 100/3, and gear 2 turns 45 times per minute, then the time when they align is LCM of 100/3 and 45, divided by something? Wait, but LCM is usually for integers. So, if you have two numbers, say, 100/3 and 45, how do you find LCM?

But since LCM is defined for integers, perhaps we need to convert them to integers by multiplying by a common factor. For example, if we consider the number of rotations, but that might not be straightforward.

Alternatively, think in terms of how many minutes it takes for both gears to complete an integer number of rotations. So, for gear 1, the number of minutes needed to complete an integer number of rotations is t = k / (100/3) = 3k/100, where k is integer. Similarly, for gear 2, t = m / 45, where m is integer. So, we need 3k/100 = m / 45. Which is the same equation as before. So, 3k/100 = m/45 => 135k = 100m => 27k = 20m. Which again leads to k=20n, m=27n, so t= 3*20n /100 = 60n /100 = 3n/5. Therefore, minimal t is 3/5 minutes, which is 36 seconds. So, same result.

Therefore, regardless of the approach, the answer is 36 seconds. Therefore, I think that is correct.

But just to make sure, let me think of another way. Let me think about the time it takes for each gear to point north again.

For gear 1: since it makes 100/3 rotations per minute, the time between each mark pointing north is 1 / (100/3) = 3/100 minutes. So, every 3/100 minutes, the mark points north again. Similarly, for gear 2, it's 1/45 minutes. So, the question is when will both 3/100 and 1/45 divide t? So, t must be a common multiple of 3/100 and 1/45. Which again is LCM(3/100, 1/45) = 3/5 minutes. So, same answer.

Alternatively, think of the problem as a clock problem. Imagine two clocks, one that ticks every 3/100 minutes, and another every 1/45 minutes. When will they both tick at the same time? Which is LCM of their periods. So, same thing.

Alternatively, think in terms of angular speed. The angular speed of gear 1 is ω1 = 2π * (100/3) radians per minute. Similarly, ω2 = 2π * 45 radians per minute. The marks will point north again when each gear has rotated an integer multiple of 2π radians. So, the time t when both gears have rotated integer multiples of 2π radians. Therefore:

ω1 * t = 2π * k => t = (2π * k)/ω1 = (2π * k) / (2π * 100/3) ) = (k * 3)/100 minutes.

Similarly, ω2 * t = 2π * m => t = (2π * m)/ω2 = (2π * m)/(2π * 45) ) = m / 45 minutes.

Therefore, same equations as before. So, t must be equal to 3k/100 and m/45. Therefore, same result. So, t=3/5 minutes. So, 36 seconds.

Therefore, all approaches lead to 36 seconds. Therefore, I think that is the correct answer.

But just to make sure, let me check with actual time.

First gear: 33 1/3 rotations per minute. So, in 36 seconds, which is 0.6 minutes, how many rotations?

0.6 minutes * 100/3 rotations per minute = 0.6 * 100 / 3 = 60 / 3 = 20 rotations. So, 20 full rotations. So, mark points north again.

Second gear: 45 rotations per minute. In 0.6 minutes, 45 * 0.6 = 27 rotations. Also integer. So, mark points north again. Perfect.

If I check 36 seconds, which is 0.6 minutes, both gears have completed integer number of rotations. Therefore, their marks point north again. And since there is no smaller time (as checked earlier), 36 seconds is the minimal time. Therefore, answer is 36 seconds.

Therefore, I think the answer is 36 seconds.

**Final Answer**
The two gears will next have both their marks pointing due north after \boxed{36} seconds.

To determine when both gears will have their marks pointing due north again, we analyze the rotational behavior of each gear and find the **least common multiple (LCM)** of their **periods**—the time it takes for each gear to complete one full rotation.

---

### Step 1: Convert Rotational Rates to Periods

- **Gear 1** rotates at $ 33\frac{1}{3} $ rotations per minute.
  Converting to an improper fraction:
  $$
  33\frac{1}{3} = \frac{100}{3} \text{ rotations per minute}
  $$
  So, the **period** (time per rotation) is:
  $$
  \frac{1}{\frac{100}{3}} = \frac{3}{100} \text{ minutes per rotation}
  $$

- **Gear 2** rotates at 45 rotations per minute.
  So, its **period** is:
  $$
  \frac{1}{45} \text{ minutes per rotation}
  $$

---

### Step 2: Find the LCM of the Two Periods

We are looking for the **smallest time** $ t $ such that both gears have completed an integer number of rotations. That is, $ t $ must be a **common multiple** of $ \frac{3}{100} $ and $ \frac{1}{45} $.

To find the **LCM of two fractions**, we use the formula:

$$
\text{LCM}\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a, c)}{\text{GCD}(b, d)}
$$

Apply this to our values:

- Numerators: $ a = 3 $, $ c = 1 $ → $ \text{LCM}(3, 1) = 3 $
- Denominators: $ b = 100 $, $ d = 45 $ → $ \text{GCD}(100, 45) = 5 $

So,
$$
\text{LCM}\left(\frac{3}{100}, \frac{1}{45}\right) = \frac{3}{5} \text{ minutes}
$$

---

### Step 3: Convert to Seconds

$$
\frac{3}{5} \text{ minutes} = \frac{3}{5} \times 60 = 36 \text{ seconds}
$$

---

### Step 4: Verification

- **Gear 1** in 36 seconds:
  $$
  \frac{100}{3} \text{ rotations per minute} \times \frac{36}{60} = \frac{100}{3} \times \frac{3}{5} = 20 \text{ rotations}
  $$

- **Gear 2** in 36 seconds:
  $$
  45 \text{ rotations per minute} \times \frac{36}{60} = 45 \times \frac{3}{5} = 27 \text{ rotations}
  $$

Both are integers, so the marks point due north again at this time.

---

### Final Answer

$$
\boxed{36}
$$

[Assistant]
Explanation:
Okay, so I need to figure out after how many seconds both gears will have their marks pointing due north again. Let me start by understanding the problem.

First, there are two gears. One turns 33 and 1/3 times per minute, and the other turns 45 times per minute. Initially, both have a mark pointing due north. I need to find the time when both marks will point north again simultaneously. The answer should be in seconds.

Hmm, gears... when gears are meshed together, their rotations are related. But wait, the problem doesn't mention them being connected. Wait, but maybe they are separate gears? Or maybe they are connected? Wait, the problem says "a mark on each gear is pointing due north." So maybe they are separate gears, not connected? But then, how does their rotation affect each other? Wait, maybe they are connected? Wait, the problem doesn't specify. Hmm. Wait, but maybe it's not necessary? Maybe it's just about the rotation of each gear independently?

Wait, but if they are separate gears, then their marks pointing north would depend on their own rotation. So, for each gear, the mark will point north every time it completes a full rotation. So, for the first gear, which turns 33 1/3 times per minute, how long does it take for the mark to point north again? Similarly for the second gear. Then, the question is when both will have their marks pointing north again, which would be the least common multiple (LCM) of their individual periods.

But let me think again. If they are separate gears, then their marks will point north again after their individual periods. So, the first gear's period is 1 / (33 1/3) minutes per rotation, and the second gear's period is 1 / 45 minutes per rotation. Then, the time when both marks point north again is the LCM of these two periods. But since LCM is usually for integers, maybe I need to convert them into fractions?

Alternatively, maybe think in terms of how many rotations each gear makes in a certain time, and find when both have completed an integer number of rotations. So, the time when both marks point north again is the smallest time t such that t multiplied by the number of rotations per minute for each gear is an integer. So, for the first gear: t * 33 1/3 must be integer, and for the second gear: t * 45 must be integer. Then, find the smallest t (in minutes) that satisfies both, then convert to seconds.

Yes, that seems like a better approach. Let me formalize this.

Let me denote:

For gear 1: rotation rate is 33 1/3 rotations per minute. Let me convert that to an improper fraction. 33 1/3 is equal to 100/3 rotations per minute. So, 100/3 rpm.

For gear 2: 45 rotations per minute.

So, if I let t be the time in minutes, then the number of rotations for gear 1 is (100/3) * t, and for gear 2 is 45 * t. We need both of these numbers to be integers. So, (100/3)*t must be integer, and 45*t must be integer.

So, let me write equations:

(100/3)*t ∈ ℤ

and

45*t ∈ ℤ

So, t must be such that t is a multiple of 3/100 minutes to make (100/3)*t integer. Wait, let me think. Let me solve for t.

For the first equation: (100/3)*t is integer. Let me denote that as:

Let me write t = k / (100/3) where k is integer? Wait, no. Let me think again. If (100/3)*t is integer, then t must be a multiple of 3/100. Because if you have (100/3)*t = integer, then t = (integer) * 3/100. Similarly, for the second equation, 45*t is integer, so t must be a multiple of 1/45.

Therefore, t must be a common multiple of 3/100 and 1/45. Therefore, the least common multiple of these two fractions. How do I compute LCM of fractions?

I remember that LCM of fractions is LCM(numerators)/GCD(denominators). Wait, or is it LCM(numerators)/GCD(denominators)? Wait, actually, the formula for LCM of fractions is LCM of numerators divided by GCD of denominators. Wait, let me check.

Alternatively, another way: Let me convert the fractions to have the same denominator. Let me think. Let me recall that LCM(a/b, c/d) = LCM(a, c) / GCD(b, d). Wait, not sure. Maybe better to think in terms of converting to decimals?

Alternatively, think of t as a multiple of both 3/100 and 1/45. So, t must be a common multiple of 3/100 and 1/45. So, to find the least such t, we can find LCM of 3/100 and 1/45. How?

Alternatively, think of t as a number such that t = (3/100)*k = (1/45)*m, where k and m are integers. So, solving for t, we can write:

(3/100)*k = (1/45)*m => m = (3/100)*k * 45 = (135/100)*k = (27/20)*k

Since m must be integer, then (27/20)*k must be integer. Therefore, k must be a multiple of 20. Let me denote k = 20*n, where n is integer. Then, m = (27/20)*(20*n) = 27*n. Therefore, t = (3/100)*k = (3/100)*(20*n) = (60/100)*n = (3/5)*n. Therefore, the minimal positive t is when n=1, so t= 3/5 minutes. Wait, but let me check if this is correct.

Wait, so according to this, t= 3/5 minutes. Let me check if this satisfies both conditions.

First, for gear 1: (100/3)*t = (100/3)*(3/5) = 100/5 = 20, which is integer. Good.

For gear 2: 45*t = 45*(3/5) = 27, which is integer. Good. So, t= 3/5 minutes is the minimal time? Wait, but is there a smaller t?

Wait, let me think again. Because if I take LCM of 3/100 and 1/45, maybe there is a smaller t. Let me try another approach.

Let me convert both periods into fractions. The period for gear 1 is 1 / (100/3) = 3/100 minutes per rotation. The period for gear 2 is 1 / 45 minutes per rotation. So, the question is, when will both gears have completed an integer number of rotations? So, the time t must be a multiple of both 3/100 and 1/45. Therefore, t is the LCM of 3/100 and 1/45.

To compute LCM of two fractions, the formula is LCM(numerator1, numerator2) / GCD(denominator1, denominator2). Wait, but I need to confirm.

Alternatively, LCM(a, b) = |a*b| / GCD(a, b). But that is for integers. For fractions, perhaps?

Alternatively, think of LCM of two numbers as the smallest number that is a multiple of both. So, if I have two numbers, say, 3/100 and 1/45, then LCM is the smallest t such that t is a multiple of both. So, t = (3/100)*k = (1/45)*m. Which is the same as before.

Alternatively, convert them to have the same denominator. Let me find a common denominator for 3/100 and 1/45. The denominators are 100 and 45. The LCM of 100 and 45 is... Let me compute LCM(100, 45). Prime factors of 100: 2^2 * 5^2. Prime factors of 45: 3^2 * 5. So LCM is 2^2 * 3^2 * 5^2 = 4 * 9 * 25 = 900. So, common denominator is 900.

So, 3/100 = (3 * 9)/900 = 27/900

1/45 = (20)/900

So, now, LCM of 27/900 and 20/900 is LCM(27, 20)/900. Since denominators are same. LCM(27, 20). 27 is 3^3, 20 is 2^2 * 5. So LCM is 2^2 * 3^3 * 5 = 4 * 27 * 5 = 4 * 135 = 540. Therefore, LCM is 540 / 900 = 3/5 minutes. Which matches the previous result. So, t= 3/5 minutes. Therefore, converting to seconds, 3/5 minutes is 3/5 * 60 seconds = 36 seconds. Therefore, the answer is 36 seconds?

Wait, but let me check again. Let me verify with another approach.

Alternatively, think in terms of how many rotations each gear makes in t seconds. Wait, but maybe better to think in terms of how many rotations per second?

Wait, let me try again. Let me convert the rotation rates to rotations per second, so that I can work in seconds.

First gear: 33 1/3 rotations per minute. 33 1/3 is 100/3 rotations per minute. So, per second, that is (100/3)/60 = 100/(3*60) = 100/180 = 5/9 rotations per second.

Second gear: 45 rotations per minute, which is 45/60 = 3/4 rotations per second.

So, the first gear makes 5/9 rotations per second, the second makes 3/4 rotations per second.

We need to find the smallest t (in seconds) such that both gears have completed an integer number of rotations. So:

For gear 1: (5/9)*t must be integer.

For gear 2: (3/4)*t must be integer.

So, solving for t:

Let me denote t must satisfy:

(5/9)*t ∈ ℤ and (3/4)*t ∈ ℤ.

So, t must be such that t is a multiple of 9/5 and 4/3. Wait, similar to before. So, t must be a common multiple of 9/5 and 4/3. Therefore, LCM of 9/5 and 4/3.

Again, using the same method as before. Let me convert them to fractions with common denominators or use the formula.

Alternatively, find t such that t = (9/5)*k = (4/3)*m, where k and m are integers. Then:

(9/5)*k = (4/3)*m => 27k = 20m => 27k = 20m. Therefore, 27 divides 20m. Since GCD(27,20)=1, 27 divides m. Let m=27n. Then, 27k = 20*27n => k=20n. Therefore, t = (9/5)*k = (9/5)*20n = 36n. Therefore, the minimal t is 36 seconds when n=1. Therefore, the answer is 36 seconds. Which matches the previous result. So, that seems consistent.

Therefore, the answer is 36 seconds. Let me check once more with actual numbers.

First gear: 5/9 rotations per second. After 36 seconds, number of rotations is 5/9 * 36 = 20 rotations. Which is integer. Good.

Second gear: 3/4 rotations per second. After 36 seconds, number of rotations is 3/4 * 36 = 27 rotations. Which is integer. Perfect. So, both gears have completed integer number of rotations, so their marks point north again. And since 36 is the minimal such time, that's the answer.

But let me check if there is a smaller time. Suppose, for example, 18 seconds. Let's check:

First gear: 5/9 * 18 = 10 rotations. Integer. Second gear: 3/4 * 18 = 13.5 rotations. Not integer. So, no. How about 24 seconds?

First gear: 5/9 *24 = 13.333... Not integer. Second gear: 3/4 *24 = 18. So, only second gear is integer. Not both. How about 12 seconds?

First gear: 5/9 *12 = 6.666... Not integer. Second gear: 3/4 *12 = 9. Not both. How about 36 seconds? As above, both integers. What about 18 seconds? As above, only second gear. How about 9 seconds?

First gear: 5/9 *9 = 5. Second gear: 3/4 *9 = 6.75. Not integer. So, no. How about 6 seconds?

First gear: 5/9 *6 = 10/3 ≈ 3.333. Not integer. Second gear: 3/4 *6 = 4.5. Not integer. How about 27 seconds?

First gear: 5/9 *27 = 15. Second gear: 3/4 *27 = 20.25. Not integer. How about 36 seconds? As before, both integers. So, seems like 36 is the minimal. Therefore, answer is 36 seconds.

Alternatively, another approach: find the LCM of the periods. The period of gear 1 is 3/100 minutes, period of gear 2 is 1/45 minutes. So, LCM(3/100, 1/45). As we calculated before, LCM is 3/5 minutes, which is 36 seconds. So, same answer.

Alternatively, think in terms of how many minutes it takes for both gears to align. Since gear 1 makes 100/3 rotations per minute, gear 2 makes 45 rotations per minute. The time when both marks point north again is when both have made an integer number of rotations. So, the number of rotations for gear 1 is (100/3)*t, and for gear 2 is 45*t. So, we need (100/3)*t and 45*t to be integers. Let me write equations:

Let me denote t as minutes. So:

(100/3)*t = integer => t = (3/100)*integer.

Similarly, 45*t = integer => t = (1/45)*integer.

Therefore, t must be a common multiple of 3/100 and 1/45. So, the LCM of these two fractions. As before, LCM(3/100, 1/45) = LCM(3,1)/GCD(100,45) ? Wait, no. Wait, earlier we had LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). Wait, let me check.

Wait, if I have two fractions, say, a/b and c/d, then LCM(a/b, c/d) is LCM(a,c)/GCD(b,d). Wait, but I need to verify.

Alternatively, think of LCM of two fractions as the smallest positive number that is a multiple of both fractions. So, if I have two numbers, say, 3/100 and 1/45, then LCM is the smallest t such that t = 3/100 * k = 1/45 * m, where k and m are integers. Which is the same as before. Which led us to t=3/5 minutes. So, 3/5 minutes is 36 seconds.

Alternatively, think of the problem as finding t such that t is a multiple of both periods. So, the periods are 3/100 minutes and 1/45 minutes. So, LCM(3/100, 1/45). To compute LCM of two numbers, you can use the formula LCM(a,b) = |a*b| / GCD(a,b). But this formula is for integers. For fractions, perhaps?

Wait, if I consider the periods as numbers, then LCM(a,b) = LCM(numerators)/GCD(denominators) ?

Wait, let me check with the previous numbers. For 3/100 and 1/45. If I take LCM of numerators 3 and 1, which is 3. GCD of denominators 100 and 45 is 5. Therefore, LCM would be 3 / 5 = 3/5. Which is exactly what we had. So, that seems to be the formula. Therefore, LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). Therefore, in this case, LCM(3,1)/GCD(100,45) = 3 / 5. Therefore, LCM is 3/5 minutes. Which is 36 seconds. So, that's another way to get the same answer.

Therefore, regardless of the method, the answer is 36 seconds. Therefore, I think that is the correct answer.

But just to make sure, let me think if there's another way someone might approach this problem and maybe get confused.

For example, someone might think that since gear 1 turns 33 1/3 times per minute, which is 100/3, and gear 2 turns 45 times per minute, then the time when they align is LCM of 100/3 and 45, divided by something? Wait, but LCM is usually for integers. So, if you have two numbers, say, 100/3 and 45, how do you find LCM?

But since LCM is defined for integers, perhaps we need to convert them to integers by multiplying by a common factor. For example, if we consider the number of rotations, but that might not be straightforward.

Alternatively, think in terms of how many minutes it takes for both gears to complete an integer number of rotations. So, for gear 1, the number of minutes needed to complete an integer number of rotations is t = k / (100/3) = 3k/100, where k is integer. Similarly, for gear 2, t = m / 45, where m is integer. So, we need 3k/100 = m / 45. Which is the same equation as before. So, 3k/100 = m/45 => 135k = 100m => 27k = 20m. Which again leads to k=20n, m=27n, so t= 3*20n /100 = 60n /100 = 3n/5. Therefore, minimal t is 3/5 minutes, which is 36 seconds. So, same result.

Therefore, regardless of the approach, the answer is 36 seconds. Therefore, I think that is correct.

But just to make sure, let me think of another way. Let me think about the time it takes for each gear to point north again.

For gear 1: since it makes 100/3 rotations per minute, the time between each mark pointing north is 1 / (100/3) = 3/100 minutes. So, every 3/100 minutes, the mark points north again. Similarly, for gear 2, it's 1/45 minutes. So, the question is when will both 3/100 and 1/45 divide t? So, t must be a common multiple of 3/100 and 1/45. Which again is LCM(3/100, 1/45) = 3/5 minutes. So, same answer.

Alternatively, think of the problem as a clock problem. Imagine two clocks, one that ticks every 3/100 minutes, and another every 1/45 minutes. When will they both tick at the same time? Which is LCM of their periods. So, same thing.

Alternatively, think in terms of angular speed. The angular speed of gear 1 is ω1 = 2π * (100/3) radians per minute. Similarly, ω2 = 2π * 45 radians per minute. The marks will point north again when each gear has rotated an integer multiple of 2π radians. So, the time t when both gears have rotated integer multiples of 2π radians. Therefore:

ω1 * t = 2π * k => t = (2π * k)/ω1 = (2π * k) / (2π * 100/3) ) = (k * 3)/100 minutes.

Similarly, ω2 * t = 2π * m => t = (2π * m)/ω2 = (2π * m)/(2π * 45) ) = m / 45 minutes.

Therefore, same equations as before. So, t must be equal to 3k/100 and m/45. Therefore, same result. So, t=3/5 minutes. So, 36 seconds.

Therefore, all approaches lead to 36 seconds. Therefore, I think that is the correct answer.

But just to make sure, let me check with actual time.

First gear: 33 1/3 rotations per minute. So, in 36 seconds, which is 0.6 minutes, how many rotations?

0.6 minutes * 100/3 rotations per minute = 0.6 * 100 / 3 = 60 / 3 = 20 rotations. So, 20 full rotations. So, mark points north again.

Second gear: 45 rotations per minute. In 0.6 minutes, 45 * 0.6 = 27 rotations. Also integer. So, mark points north again. Perfect.

If I check 36 seconds, which is 0.6 minutes, both gears have completed integer number of rotations. Therefore, their marks point north again. And since there is no smaller time (as checked earlier), 36 seconds is the minimal time. Therefore, answer is 36 seconds.

Therefore, I think the answer is 36 seconds.

**Final Answer**
The two gears will next have both their marks pointing due north after \boxed{36} seconds.

To determine when both gears will have their marks pointing due north again, we analyze the rotational behavior of each gear and find the **least common multiple (LCM)** of their **periods**—the time it takes for each gear to complete one full rotation.

---

### Step 1: Convert Rotational Rates to Periods

- **Gear 1** rotates at $ 33\frac{1}{3} $ rotations per minute.
  Converting to an improper fraction:
  $$
  33\frac{1}{3} = \frac{100}{3} \text{ rotations per minute}
  $$
  So, the **period** (time per rotation) is:
  $$
  \frac{1}{\frac{100}{3}} = \frac{3}{100} \text{ minutes per rotation}
  $$

- **Gear 2** rotates at 45 rotations per minute.
  So, its **period** is:
  $$
  \frac{1}{45} \text{ minutes per rotation}
  $$

---

### Step 2: Find the LCM of the Two Periods

We are looking for the **smallest time** $ t $ such that both gears have completed an integer number of rotations. That is, $ t $ must be a **common multiple** of $ \frac{3}{100} $ and $ \frac{1}{45} $.

To find the **LCM of two fractions**, we use the formula:

$$
\text{LCM}\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a, c)}{\text{GCD}(b, d)}
$$

Apply this to our values:

- Numerators: $ a = 3 $, $ c = 1 $ → $ \text{LCM}(3, 1) = 3 $
- Denominators: $ b = 100 $, $ d = 45 $ → $ \text{GCD}(100, 45) = 5 $

So,
$$
\text{LCM}\left(\frac{3}{100}, \frac{1}{45}\right) = \frac{3}{5} \text{ minutes}
$$

---

### Step 3: Convert to Seconds

$$
\frac{3}{5} \text{ minutes} = \frac{3}{5} \times 60 = 36 \text{ seconds}
$$

---

### Step 4: Verification

- **Gear 1** in 36 seconds:
  $$
  \frac{100}{3} \text{ rotations per minute} \times \frac{36}{60} = \frac{100}{3} \times \frac{3}{5} = 20 \text{ rotations}
  $$

- **Gear 2** in 36 seconds:
  $$
  45 \text{ rotations per minute} \times \frac{36}{60} = 45 \times \frac{3}{5} = 27 \text{ rotations}
  $$

Both are integers, so the marks point due north again at this time.

---

### Final Answer

$$
\boxed{36}
$$

Solution: One gear turns $33\frac{1}{3}=100/3$ times in 60 seconds, so it turns 5/9 times in one second, or 5 times in 9 seconds. The other gear turns 45 times in 60 seconds, so it turns 3/4 times in one second, or 3 times in 4 seconds. To find out after how many seconds the two gears next have both their marks pointing due north, we have to find the least common multiple of $4=2^2$ and $9=3^2$, which is $2^2\cdot3^2=36$. Therefore, the two gears next have both their marks pointing due north after $\boxed{36}$ seconds. (One gear turns exactly $5\times4=20$ times, and the other gear turns exactly $3\times9=27$ times.)


Explanation:
Okay, so I need to find the distance between two parallel lines. Let me recall how to do that. I think when two lines are parallel, the distance between them can be found by taking any point on one line and finding the distance from that point to the other line. But since they are parallel, this distance should be the same no matter which point I choose on the first line. Alternatively, there's a formula involving the cross product of direction vectors and the vector connecting a point on each line. Wait, let me think.

First, let me write down the equations of the lines again to make sure I have them right.

First line:
$$
\begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}
$$
Second line:
$$
\begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} + u \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}
$$
They are parallel, so their direction vectors should be scalar multiples of each other. Let me check that. The direction vector of the first line is $\begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$, and the direction vector of the second line is $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$. Let me see if one is a scalar multiple of the other. If I multiply the first direction vector by 2, I get $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$, which is exactly the direction vector of the second line. So yes, they are parallel, as given. Good.

Now, to find the distance between two parallel lines. I remember that the distance can be calculated by taking the length of the cross product of the direction vector and the vector connecting a point on each line, divided by the magnitude of the direction vector. Wait, let me recall the exact formula.

The formula for the distance between two parallel lines is:

$$
\text{Distance} = \frac{|\vec{v} \times \vec{w}|}{|\vec{v}|}
$$

where $\vec{v}$ is the direction vector of the lines, and $\vec{w}$ is the vector connecting any point on the first line to any point on the second line. Is that right? Let me verify.

Alternatively, since the lines are parallel, the distance between them is the length of the component of the vector connecting the two points that is perpendicular to the direction vector. So if I take a vector between a point on the first line and a point on the second line, then project it onto a direction perpendicular to the common direction vector. The length of that projection is the distance between the lines.

Yes, so if I have two points, say $P$ on the first line and $Q$ on the second line, then the vector $\vec{PQ}$ connects them. The distance between the lines is the length of the projection of $\vec{PQ}$ onto a direction perpendicular to the direction vector of the lines. Since the direction vector is $\vec{v}$, then the distance is $|\vec{PQ} \times \vec{v}| / |\vec{v}|$. Wait, that seems similar to the formula I mentioned earlier. Let me check.

Alternatively, since the cross product $\vec{PQ} \times \vec{v}$ gives a vector whose magnitude is $|\vec{PQ}||\vec{v}|\sin\theta$, where $\theta$ is the angle between them. Then dividing by $|\vec{v}|$ gives $|\vec{PQ}|\sin\theta$, which is the length of the projection of $\vec{PQ}$ onto a direction perpendicular to $\vec{v}$. Which is exactly the distance between the lines. So yes, that formula is correct.

Therefore, the steps are:

1. Choose a point on each line. Let's take the points given in the parametric equations. For the first line, when $t = 0$, the point is $P = \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}$. For the second line, when $u = 0$, the point is $Q = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}$.

2. Compute the vector $\vec{PQ} = Q - P = \begin{pmatrix} -5 - 3 \\ -3 - (-10) \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$.

3. Take the direction vector of the lines, which is $\vec{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$ (or the other direction vector, but since we are taking cross product, it doesn't matter as cross product is anti-commutative, but magnitude will be same).

4. Compute the cross product $\vec{PQ} \times \vec{v}$.

5. Find the magnitude of that cross product, then divide by the magnitude of $\vec{v}$ to get the distance.

Let me compute that step by step.

First, compute $\vec{PQ} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$, and $\vec{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$.

Compute cross product $\vec{PQ} \times \vec{v}$:

The cross product of vectors $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and $\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$ is:

$$
\begin{pmatrix}
a_2b_3 - a_3b_2 \\
a_3b_1 - a_1b_3 \\
a_1b_2 - a_2b_1
\end{pmatrix}
$$

So applying that:

First component: $a_2b_3 - a_3b_2 = (7)(-2) - (5)(-9) = -14 + 45 = 31$

Second component: $a_3b_1 - a_1b_3 = (5)(2) - (-8)(-2) = 10 - 16 = -6$

Third component: $a_1b_2 - a_2b_1 = (-8)(-9) - (7)(2) = 72 - 14 = 58$

So the cross product vector is $\begin{pmatrix} 31 \\ -6 \\ 58 \end{pmatrix}$.

Now compute its magnitude:

$$
|\vec{PQ} \times \vec{v}| = \sqrt{31^2 + (-6)^2 + 58^2}
$$

Let me compute each term:

31 squared is 961.

(-6) squared is 36.

58 squared is 3364.

Adding them up: 961 + 36 = 997; 997 + 3364 = 4361.

So the magnitude is sqrt(4361). Let me check if that can be simplified or if I made a calculation error.

Wait, let me verify the cross product components again because that seems like a big number. Let me recheck:

First component: 7*(-2) - 5*(-9) = -14 + 45 = 31. Correct.

Second component: 5*2 - (-8)*(-2) = 10 - 16 = -6. Correct.

Third component: (-8)*(-9) - 7*2 = 72 - 14 = 58. Correct. So cross product is correct.

So magnitude squared is 31² + (-6)² + 58² = 961 + 36 + 3364. Let me add again:

961 + 36 = 997. 997 + 3364: 997 + 3000 = 3997, 3997 + 364 = 4361. Yes, that's correct.

So sqrt(4361). Let me see if that's a perfect square or can be simplified. Let me check sqrt(4361). Let me see, 66² is 4356, 67² is 4489. So sqrt(4361) is between 66 and 67. 66² = 4356, so 4361 - 4356 = 5. So sqrt(4361) = sqrt(66² + 5). Not a perfect square. So maybe I need to keep it as sqrt(4361) for now.

Now, compute the magnitude of direction vector $\vec{v}$:

$$
|\vec{v}| = \sqrt{2^2 + (-9)^2 + (-2)^2} = \sqrt{4 + 81 + 4} = \sqrt{89}
$$

Therefore, the distance is sqrt(4361)/sqrt(89). Let me simplify that.

Note that sqrt(a)/sqrt(b) = sqrt(a/b). So sqrt(4361 / 89). Let me compute 4361 divided by 89.

Let me do the division: 89 * 49 = 89*50 - 89 = 4450 - 89 = 4361. Oh! So 89*49 = 4361. Therefore, sqrt(4361 / 89) = sqrt(49) = 7. Wait, that's nice!

So the distance is 7. Wait, that seems too straightforward. Let me check again.

Yes, 89*49: 89*50 is 4450, minus 89 is 4450 - 89 = 4361. Correct. Therefore, sqrt(4361 / 89) = sqrt(49) = 7. Therefore, the distance between the two lines is 7. That seems surprisingly simple. Let me check if I did everything correctly.

Wait, so cross product magnitude was sqrt(4361), divided by sqrt(89) gives sqrt(4361/89) = sqrt(49) = 7. So yes, the distance is 7. That seems correct. Let me think again if there's another way to check.

Alternatively, since the lines are parallel, we can parametrize one line and find the minimum distance. Let me try that approach as a check.

Let me take the first line: P(t) = (3, -10, 1) + t(2, -9, -2)

Second line: Q(u) = (-5, -3, 6) + u(4, -18, -4). Since direction vectors are scalar multiples, as we saw earlier, direction vector is (2, -9, -2) multiplied by 2.

So, to find the distance between the lines, we can take a point on the first line, say P(t), and find the distance to the second line, then minimize it. But since the lines are parallel, the minimal distance should be constant. Wait, but maybe not? Wait, no, if they are parallel, the minimal distance is the same between any two points. Wait, but actually, if they are parallel and not coinciding, then the minimal distance is constant. So if I compute the distance from a point on the first line to the second line, that should give me the distance between the lines.

Let me try that. Let me take the point P(0) = (3, -10, 1) on the first line. Let me find the distance from this point to the second line.

The formula for the distance from a point to a line in 3D is |(Q - P) × direction| / |direction|, where Q is a point on the line. Wait, that's exactly what I did earlier. So that gives the same result. So since I already did that, and got 7, that's consistent.

Alternatively, let me try parametrizing the distance squared between P(t) and Q(u) and minimize it. Let me try that.

Let me denote P(t) = (3 + 2t, -10 -9t, 1 - 2t)

Q(u) = (-5 + 4u, -3 -18u, 6 -4u)

The vector between P(t) and Q(u) is:

( -5 + 4u - 3 - 2t, -3 -18u - (-10 -9t), 6 -4u - (1 - 2t) )

Simplify each component:

x-component: -5 -3 + 4u - 2t = -8 + 4u - 2t

y-component: -3 +10 -18u +9t = 7 -18u +9t

z-component: 6 -1 -4u +2t = 5 -4u +2t

So the distance squared between P(t) and Q(u) is:

[ -8 + 4u - 2t ]² + [ 7 -18u +9t ]² + [ 5 -4u +2t ]²

Since the lines are parallel, the minimal distance should be achieved when the vector between P(t) and Q(u) is perpendicular to the direction vector of the lines. The direction vector is (2, -9, -2). So the vector between P(t) and Q(u) should be perpendicular to (2, -9, -2). Therefore, their dot product is zero.

So, let me denote vector between P(t) and Q(u) as (A, B, C) = (-8 + 4u - 2t, 7 -18u +9t, 5 -4u +2t)

Dot product with direction vector (2, -9, -2):

2*A + (-9)*B + (-2)*C = 0

So:

2*(-8 + 4u - 2t) -9*(7 -18u +9t) -2*(5 -4u +2t) = 0

Let me compute each term:

First term: 2*(-8 + 4u - 2t) = -16 + 8u -4t

Second term: -9*(7 -18u +9t) = -63 + 162u -81t

Third term: -2*(5 -4u +2t) = -10 +8u -4t

Adding all together:

(-16 -63 -10) + (8u +162u +8u) + (-4t -81t -4t) = 0

Calculating constants: -16 -63 = -79; -79 -10 = -89

Calculating u terms: 8u +162u = 170u; 170u +8u = 178u

Calculating t terms: -4t -81t = -85t; -85t -4t = -89t

So overall equation:

-89 + 178u -89t = 0

Divide both sides by -89:

1 - 2u + t = 0 => t = 2u -1

So this gives a relationship between t and u. Now, since the minimal distance occurs when this condition is satisfied, we can substitute t = 2u -1 into the distance squared formula and then find the minimum.

But since the lines are parallel, the minimal distance should be the same regardless of u, but let me proceed.

So let me substitute t = 2u -1 into the components of the vector between P(t) and Q(u):

First, compute A = -8 + 4u - 2t = -8 + 4u - 2*(2u -1) = -8 +4u -4u +2 = (-8 +2) + (4u -4u) = -6 + 0 = -6

Similarly, compute B = 7 -18u +9t = 7 -18u +9*(2u -1) = 7 -18u +18u -9 = (7 -9) + (-18u +18u) = -2 + 0 = -2

Compute C = 5 -4u +2t = 5 -4u +2*(2u -1) = 5 -4u +4u -2 = (5 -2) + (-4u +4u) = 3 + 0 = 3

So the vector between P(t) and Q(u) is (-6, -2, 3). Therefore, the distance squared is (-6)^2 + (-2)^2 + 3^2 = 36 + 4 + 9 = 49. Therefore, the distance is sqrt(49) = 7. Which matches the previous result. So that's a good check. Therefore, the distance between the two lines is indeed 7.

So, even though the cross product calculation gave me sqrt(4361)/sqrt(89) which simplified to 7, and the parametrization method also gives 7, so I can be confident that the answer is 7.

But just to make sure, let me think if there's another way to approach this. For example, using the formula for the distance between two parallel lines in terms of a point and the direction vector.

Alternatively, since the lines are parallel, we can take a vector connecting any two points on the lines, say from P(0) to Q(0), which is (-8, 7, 5) as before. Then, the distance is the norm of the component of this vector that is perpendicular to the direction vector.

The formula for the component of vector $\vec{PQ}$ perpendicular to $\vec{v}$ is $|\vec{PQ} \times \vec{v}| / |\vec{v}|$, which is exactly what I calculated earlier. Which again gives 7.

Alternatively, we can compute the projection of $\vec{PQ}$ onto $\vec{v}$, subtract that from $\vec{PQ}$, and then take the magnitude. Let me try that.

The projection of $\vec{PQ}$ onto $\vec{v}$ is:

$$
\text{proj}_{\vec{v}} \vec{PQ} = \frac{\vec{PQ} \cdot \vec{v}}{|\vec{v}|^2} \vec{v}
$$

Then, the component of $\vec{PQ}$ perpendicular to $\vec{v}$ is $\vec{PQ} - \text{proj}_{\vec{v}} \vec{PQ}$. The magnitude of this vector is the distance between the lines.

Let me compute that.

First, compute $\vec{PQ} \cdot \vec{v}$:

$\vec{PQ} = (-8, 7, 5)$, $\vec{v} = (2, -9, -2)$

Dot product: (-8)(2) + (7)(-9) + (5)(-2) = -16 -63 -10 = -89

Then, $|\vec{v}|^2 = 2^2 + (-9)^2 + (-2)^2 = 4 + 81 + 4 = 89$

Therefore, the projection vector is:

$$
\frac{-89}{89} \vec{v} = -1 \cdot \vec{v} = \begin{pmatrix} -2 \\ 9 \\ 2 \end{pmatrix}
$$

Therefore, the perpendicular component is $\vec{PQ} - (-\vec{v}) = \vec{PQ} + \vec{v} = (-8 + 2, 7 + (-9), 5 + (-2)) = (-6, -2, 3)$. Which is the same vector as before! Then, the magnitude of this vector is sqrt((-6)^2 + (-2)^2 + 3^2) = sqrt(36 + 4 + 9) = sqrt(49) = 7. So again, same result.

Therefore, all methods lead to the distance being 7. Therefore, I can be confident that the answer is 7.

But just to recap, the key steps were:

1. Confirm that the lines are parallel by checking direction vectors are scalar multiples.

2. Choose a point on each line, compute the vector between them.

3. Use cross product formula to find the distance, or parametrize and minimize the distance, or compute the perpendicular component.

All methods led to the same answer. Therefore, the distance between the two lines is 7.

**Final Answer**
The distance between the two lines is \boxed{7}.

To find the distance between two **parallel lines** in 3D space, we can use the following geometric approach:

---

### **Step 1: Confirm the Lines are Parallel**

We are given two lines:

- **Line 1**:
  $$
  \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}
  $$
  Direction vector:
  $$
  \vec{v}_1 = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}
  $$

- **Line 2**:
  $$
  \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} + u \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}
  $$
  Direction vector:
  $$
  \vec{v}_2 = \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}
  $$

Check if $\vec{v}_2$ is a scalar multiple of $\vec{v}_1$:

$$
\vec{v}_2 = 2 \cdot \vec{v}_1 = 2 \cdot \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}
$$

✅ The lines are **parallel**.

---

### **Step 2: Choose Points on Each Line**

Let’s take:

- Point $P$ on Line 1:
  $$
  P = \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}
  $$

- Point $Q$ on Line 2:
  $$
  Q = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}
  $$

Compute the vector $\vec{PQ}$:

$$
\vec{PQ} = Q - P = \begin{pmatrix} -5 - 3 \\ -3 - (-10) \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}
$$

---

### **Step 3: Use the Cross Product Formula**

The distance $D$ between two parallel lines is given by:

$$
D = \frac{|\vec{PQ} \times \vec{v}|}{|\vec{v}|}
$$

Where:
- $\vec{v}$ is the direction vector of the lines (we can use $\vec{v}_1 = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$)
- $\vec{PQ} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$

Compute the cross product $\vec{PQ} \times \vec{v}$:

$$
\vec{PQ} \times \vec{v} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-8 & 7 & 5 \\
2 & -9 & -2
\end{vmatrix}
= \mathbf{i}(7 \cdot -2 - 5 \cdot -9) - \mathbf{j}(-8 \cdot -2 - 5 \cdot 2) + \mathbf{k}(-8 \cdot -9 - 7 \cdot 2)
$$

$$
= \mathbf{i}(-14 + 45) - \mathbf{j}(16 - 10) + \mathbf{k}(72 - 14)
= \begin{pmatrix} 31 \\ -6 \\ 58 \end{pmatrix}
$$

Now compute the magnitude of the cross product:

$$
|\vec{PQ} \times \vec{v}| = \sqrt{31^2 + (-6)^2 + 58^2} = \sqrt{961 + 36 + 3364} = \sqrt{4361}
$$

Compute the magnitude of the direction vector:

$$
|\vec{v}| = \sqrt{2^2 + (-9)^2 + (-2)^2} = \sqrt{4 + 81 + 4} = \sqrt{89}
$$

Now compute the distance:

$$
D = \frac{\sqrt{4361}}{\sqrt{89}} = \sqrt{\frac{4361}{89}} = \sqrt{49} = 7
$$

---

### **Final Answer**

$$
\boxed{7}
$$

[Assistant]
Explanation:
Okay, so I need to find the distance between two parallel lines. Let me recall how to do that. I think when two lines are parallel, the distance between them can be found by taking any point on one line and finding the distance from that point to the other line. But since they are parallel, this distance should be the same no matter which point I choose on the first line. Alternatively, there's a formula involving the cross product of direction vectors and the vector connecting a point on each line. Wait, let me think.

First, let me write down the equations of the lines again to make sure I have them right.

First line:
$$
\begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}
$$
Second line:
$$
\begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} + u \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}
$$
They are parallel, so their direction vectors should be scalar multiples of each other. Let me check that. The direction vector of the first line is $\begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$, and the direction vector of the second line is $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$. Let me see if one is a scalar multiple of the other. If I multiply the first direction vector by 2, I get $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$, which is exactly the direction vector of the second line. So yes, they are parallel, as given. Good.

Now, to find the distance between two parallel lines. I remember that the distance can be calculated by taking the length of the cross product of the direction vector and the vector connecting a point on each line, divided by the magnitude of the direction vector. Wait, let me recall the exact formula.

The formula for the distance between two parallel lines is:

$$
\text{Distance} = \frac{|\vec{v} \times \vec{w}|}{|\vec{v}|}
$$

where $\vec{v}$ is the direction vector of the lines, and $\vec{w}$ is the vector connecting any point on the first line to any point on the second line. Is that right? Let me verify.

Alternatively, since the lines are parallel, the distance between them is the length of the component of the vector connecting the two points that is perpendicular to the direction vector. So if I take a vector between a point on the first line and a point on the second line, then project it onto a direction perpendicular to the common direction vector. The length of that projection is the distance between the lines.

Yes, so if I have two points, say $P$ on the first line and $Q$ on the second line, then the vector $\vec{PQ}$ connects them. The distance between the lines is the length of the projection of $\vec{PQ}$ onto a direction perpendicular to the direction vector of the lines. Since the direction vector is $\vec{v}$, then the distance is $|\vec{PQ} \times \vec{v}| / |\vec{v}|$. Wait, that seems similar to the formula I mentioned earlier. Let me check.

Alternatively, since the cross product $\vec{PQ} \times \vec{v}$ gives a vector whose magnitude is $|\vec{PQ}||\vec{v}|\sin\theta$, where $\theta$ is the angle between them. Then dividing by $|\vec{v}|$ gives $|\vec{PQ}|\sin\theta$, which is the length of the projection of $\vec{PQ}$ onto a direction perpendicular to $\vec{v}$. Which is exactly the distance between the lines. So yes, that formula is correct.

Therefore, the steps are:

1. Choose a point on each line. Let's take the points given in the parametric equations. For the first line, when $t = 0$, the point is $P = \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}$. For the second line, when $u = 0$, the point is $Q = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}$.

2. Compute the vector $\vec{PQ} = Q - P = \begin{pmatrix} -5 - 3 \\ -3 - (-10) \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$.

3. Take the direction vector of the lines, which is $\vec{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$ (or the other direction vector, but since we are taking cross product, it doesn't matter as cross product is anti-commutative, but magnitude will be same).

4. Compute the cross product $\vec{PQ} \times \vec{v}$.

5. Find the magnitude of that cross product, then divide by the magnitude of $\vec{v}$ to get the distance.

Let me compute that step by step.

First, compute $\vec{PQ} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$, and $\vec{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$.

Compute cross product $\vec{PQ} \times \vec{v}$:

The cross product of vectors $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and $\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$ is:

$$
\begin{pmatrix}
a_2b_3 - a_3b_2 \\
a_3b_1 - a_1b_3 \\
a_1b_2 - a_2b_1
\end{pmatrix}
$$

So applying that:

First component: $a_2b_3 - a_3b_2 = (7)(-2) - (5)(-9) = -14 + 45 = 31$

Second component: $a_3b_1 - a_1b_3 = (5)(2) - (-8)(-2) = 10 - 16 = -6$

Third component: $a_1b_2 - a_2b_1 = (-8)(-9) - (7)(2) = 72 - 14 = 58$

So the cross product vector is $\begin{pmatrix} 31 \\ -6 \\ 58 \end{pmatrix}$.

Now compute its magnitude:

$$
|\vec{PQ} \times \vec{v}| = \sqrt{31^2 + (-6)^2 + 58^2}
$$

Let me compute each term:

31 squared is 961.

(-6) squared is 36.

58 squared is 3364.

Adding them up: 961 + 36 = 997; 997 + 3364 = 4361.

So the magnitude is sqrt(4361). Let me check if that can be simplified or if I made a calculation error.

Wait, let me verify the cross product components again because that seems like a big number. Let me recheck:

First component: 7*(-2) - 5*(-9) = -14 + 45 = 31. Correct.

Second component: 5*2 - (-8)*(-2) = 10 - 16 = -6. Correct.

Third component: (-8)*(-9) - 7*2 = 72 - 14 = 58. Correct. So cross product is correct.

So magnitude squared is 31² + (-6)² + 58² = 961 + 36 + 3364. Let me add again:

961 + 36 = 997. 997 + 3364: 997 + 3000 = 3997, 3997 + 364 = 4361. Yes, that's correct.

So sqrt(4361). Let me see if that's a perfect square or can be simplified. Let me check sqrt(4361). Let me see, 66² is 4356, 67² is 4489. So sqrt(4361) is between 66 and 67. 66² = 4356, so 4361 - 4356 = 5. So sqrt(4361) = sqrt(66² + 5). Not a perfect square. So maybe I need to keep it as sqrt(4361) for now.

Now, compute the magnitude of direction vector $\vec{v}$:

$$
|\vec{v}| = \sqrt{2^2 + (-9)^2 + (-2)^2} = \sqrt{4 + 81 + 4} = \sqrt{89}
$$

Therefore, the distance is sqrt(4361)/sqrt(89). Let me simplify that.

Note that sqrt(a)/sqrt(b) = sqrt(a/b). So sqrt(4361 / 89). Let me compute 4361 divided by 89.

Let me do the division: 89 * 49 = 89*50 - 89 = 4450 - 89 = 4361. Oh! So 89*49 = 4361. Therefore, sqrt(4361 / 89) = sqrt(49) = 7. Wait, that's nice!

So the distance is 7. Wait, that seems too straightforward. Let me check again.

Yes, 89*49: 89*50 is 4450, minus 89 is 4450 - 89 = 4361. Correct. Therefore, sqrt(4361 / 89) = sqrt(49) = 7. Therefore, the distance between the two lines is 7. That seems surprisingly simple. Let me check if I did everything correctly.

Wait, so cross product magnitude was sqrt(4361), divided by sqrt(89) gives sqrt(4361/89) = sqrt(49) = 7. So yes, the distance is 7. That seems correct. Let me think again if there's another way to check.

Alternatively, since the lines are parallel, we can parametrize one line and find the minimum distance. Let me try that approach as a check.

Let me take the first line: P(t) = (3, -10, 1) + t(2, -9, -2)

Second line: Q(u) = (-5, -3, 6) + u(4, -18, -4). Since direction vectors are scalar multiples, as we saw earlier, direction vector is (2, -9, -2) multiplied by 2.

So, to find the distance between the lines, we can take a point on the first line, say P(t), and find the distance to the second line, then minimize it. But since the lines are parallel, the minimal distance should be constant. Wait, but maybe not? Wait, no, if they are parallel, the minimal distance is the same between any two points. Wait, but actually, if they are parallel and not coinciding, then the minimal distance is constant. So if I compute the distance from a point on the first line to the second line, that should give me the distance between the lines.

Let me try that. Let me take the point P(0) = (3, -10, 1) on the first line. Let me find the distance from this point to the second line.

The formula for the distance from a point to a line in 3D is |(Q - P) × direction| / |direction|, where Q is a point on the line. Wait, that's exactly what I did earlier. So that gives the same result. So since I already did that, and got 7, that's consistent.

Alternatively, let me try parametrizing the distance squared between P(t) and Q(u) and minimize it. Let me try that.

Let me denote P(t) = (3 + 2t, -10 -9t, 1 - 2t)

Q(u) = (-5 + 4u, -3 -18u, 6 -4u)

The vector between P(t) and Q(u) is:

( -5 + 4u - 3 - 2t, -3 -18u - (-10 -9t), 6 -4u - (1 - 2t) )

Simplify each component:

x-component: -5 -3 + 4u - 2t = -8 + 4u - 2t

y-component: -3 +10 -18u +9t = 7 -18u +9t

z-component: 6 -1 -4u +2t = 5 -4u +2t

So the distance squared between P(t) and Q(u) is:

[ -8 + 4u - 2t ]² + [ 7 -18u +9t ]² + [ 5 -4u +2t ]²

Since the lines are parallel, the minimal distance should be achieved when the vector between P(t) and Q(u) is perpendicular to the direction vector of the lines. The direction vector is (2, -9, -2). So the vector between P(t) and Q(u) should be perpendicular to (2, -9, -2). Therefore, their dot product is zero.

So, let me denote vector between P(t) and Q(u) as (A, B, C) = (-8 + 4u - 2t, 7 -18u +9t, 5 -4u +2t)

Dot product with direction vector (2, -9, -2):

2*A + (-9)*B + (-2)*C = 0

So:

2*(-8 + 4u - 2t) -9*(7 -18u +9t) -2*(5 -4u +2t) = 0

Let me compute each term:

First term: 2*(-8 + 4u - 2t) = -16 + 8u -4t

Second term: -9*(7 -18u +9t) = -63 + 162u -81t

Third term: -2*(5 -4u +2t) = -10 +8u -4t

Adding all together:

(-16 -63 -10) + (8u +162u +8u) + (-4t -81t -4t) = 0

Calculating constants: -16 -63 = -79; -79 -10 = -89

Calculating u terms: 8u +162u = 170u; 170u +8u = 178u

Calculating t terms: -4t -81t = -85t; -85t -4t = -89t

So overall equation:

-89 + 178u -89t = 0

Divide both sides by -89:

1 - 2u + t = 0 => t = 2u -1

So this gives a relationship between t and u. Now, since the minimal distance occurs when this condition is satisfied, we can substitute t = 2u -1 into the distance squared formula and then find the minimum.

But since the lines are parallel, the minimal distance should be the same regardless of u, but let me proceed.

So let me substitute t = 2u -1 into the components of the vector between P(t) and Q(u):

First, compute A = -8 + 4u - 2t = -8 + 4u - 2*(2u -1) = -8 +4u -4u +2 = (-8 +2) + (4u -4u) = -6 + 0 = -6

Similarly, compute B = 7 -18u +9t = 7 -18u +9*(2u -1) = 7 -18u +18u -9 = (7 -9) + (-18u +18u) = -2 + 0 = -2

Compute C = 5 -4u +2t = 5 -4u +2*(2u -1) = 5 -4u +4u -2 = (5 -2) + (-4u +4u) = 3 + 0 = 3

So the vector between P(t) and Q(u) is (-6, -2, 3). Therefore, the distance squared is (-6)^2 + (-2)^2 + 3^2 = 36 + 4 + 9 = 49. Therefore, the distance is sqrt(49) = 7. Which matches the previous result. So that's a good check. Therefore, the distance between the two lines is indeed 7.

So, even though the cross product calculation gave me sqrt(4361)/sqrt(89) which simplified to 7, and the parametrization method also gives 7, so I can be confident that the answer is 7.

But just to make sure, let me think if there's another way to approach this. For example, using the formula for the distance between two parallel lines in terms of a point and the direction vector.

Alternatively, since the lines are parallel, we can take a vector connecting any two points on the lines, say from P(0) to Q(0), which is (-8, 7, 5) as before. Then, the distance is the norm of the component of this vector that is perpendicular to the direction vector.

The formula for the component of vector $\vec{PQ}$ perpendicular to $\vec{v}$ is $|\vec{PQ} \times \vec{v}| / |\vec{v}|$, which is exactly what I calculated earlier. Which again gives 7.

Alternatively, we can compute the projection of $\vec{PQ}$ onto $\vec{v}$, subtract that from $\vec{PQ}$, and then take the magnitude. Let me try that.

The projection of $\vec{PQ}$ onto $\vec{v}$ is:

$$
\text{proj}_{\vec{v}} \vec{PQ} = \frac{\vec{PQ} \cdot \vec{v}}{|\vec{v}|^2} \vec{v}
$$

Then, the component of $\vec{PQ}$ perpendicular to $\vec{v}$ is $\vec{PQ} - \text{proj}_{\vec{v}} \vec{PQ}$. The magnitude of this vector is the distance between the lines.

Let me compute that.

First, compute $\vec{PQ} \cdot \vec{v}$:

$\vec{PQ} = (-8, 7, 5)$, $\vec{v} = (2, -9, -2)$

Dot product: (-8)(2) + (7)(-9) + (5)(-2) = -16 -63 -10 = -89

Then, $|\vec{v}|^2 = 2^2 + (-9)^2 + (-2)^2 = 4 + 81 + 4 = 89$

Therefore, the projection vector is:

$$
\frac{-89}{89} \vec{v} = -1 \cdot \vec{v} = \begin{pmatrix} -2 \\ 9 \\ 2 \end{pmatrix}
$$

Therefore, the perpendicular component is $\vec{PQ} - (-\vec{v}) = \vec{PQ} + \vec{v} = (-8 + 2, 7 + (-9), 5 + (-2)) = (-6, -2, 3)$. Which is the same vector as before! Then, the magnitude of this vector is sqrt((-6)^2 + (-2)^2 + 3^2) = sqrt(36 + 4 + 9) = sqrt(49) = 7. So again, same result.

Therefore, all methods lead to the distance being 7. Therefore, I can be confident that the answer is 7.

But just to recap, the key steps were:

1. Confirm that the lines are parallel by checking direction vectors are scalar multiples.

2. Choose a point on each line, compute the vector between them.

3. Use cross product formula to find the distance, or parametrize and minimize the distance, or compute the perpendicular component.

All methods led to the same answer. Therefore, the distance between the two lines is 7.

**Final Answer**
The distance between the two lines is \boxed{7}.

To find the distance between two **parallel lines** in 3D space, we can use the following geometric approach:

---

### **Step 1: Confirm the Lines are Parallel**

We are given two lines:

- **Line 1**:
  $$
  \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}
  $$
  Direction vector:
  $$
  \vec{v}_1 = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}
  $$

- **Line 2**:
  $$
  \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} + u \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}
  $$
  Direction vector:
  $$
  \vec{v}_2 = \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}
  $$

Check if $\vec{v}_2$ is a scalar multiple of $\vec{v}_1$:

$$
\vec{v}_2 = 2 \cdot \vec{v}_1 = 2 \cdot \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}
$$

✅ The lines are **parallel**.

---

### **Step 2: Choose Points on Each Line**

Let’s take:

- Point $P$ on Line 1:
  $$
  P = \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}
  $$

- Point $Q$ on Line 2:
  $$
  Q = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}
  $$

Compute the vector $\vec{PQ}$:

$$
\vec{PQ} = Q - P = \begin{pmatrix} -5 - 3 \\ -3 - (-10) \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}
$$

---

### **Step 3: Use the Cross Product Formula**

The distance $D$ between two parallel lines is given by:

$$
D = \frac{|\vec{PQ} \times \vec{v}|}{|\vec{v}|}
$$

Where:
- $\vec{v}$ is the direction vector of the lines (we can use $\vec{v}_1 = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$)
- $\vec{PQ} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$

Compute the cross product $\vec{PQ} \times \vec{v}$:

$$
\vec{PQ} \times \vec{v} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-8 & 7 & 5 \\
2 & -9 & -2
\end{vmatrix}
= \mathbf{i}(7 \cdot -2 - 5 \cdot -9) - \mathbf{j}(-8 \cdot -2 - 5 \cdot 2) + \mathbf{k}(-8 \cdot -9 - 7 \cdot 2)
$$

$$
= \mathbf{i}(-14 + 45) - \mathbf{j}(16 - 10) + \mathbf{k}(72 - 14)
= \begin{pmatrix} 31 \\ -6 \\ 58 \end{pmatrix}
$$

Now compute the magnitude of the cross product:

$$
|\vec{PQ} \times \vec{v}| = \sqrt{31^2 + (-6)^2 + 58^2} = \sqrt{961 + 36 + 3364} = \sqrt{4361}
$$

Compute the magnitude of the direction vector:

$$
|\vec{v}| = \sqrt{2^2 + (-9)^2 + (-2)^2} = \sqrt{4 + 81 + 4} = \sqrt{89}
$$

Now compute the distance:

$$
D = \frac{\sqrt{4361}}{\sqrt{89}} = \sqrt{\frac{4361}{89}} = \sqrt{49} = 7
$$

---

### **Final Answer**

$$
\boxed{7}
$$

Solution:
We see that $(3,-10,1)$ is a point on the first line.

A point on the second line is given by
\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} + t \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix} = \begin{pmatrix} -5 + 4t \\ -3 - 18t \\ 6 - 4t \end{pmatrix}.\][asy]
unitsize (0.6 cm);

pair A, B, C, D, E, F, H;

A = (2,5);
B = (0,0);
C = (8,0);
D = (A + reflect(B,C)*(A))/2;

draw(A--D);
draw((0,5)--(8,5));
draw((0,0)--(8,0));

dot("$(3,-10,1)$", A, N);
dot("$(-5 + 4t, -3 - 18t, 6 - 4t)$", D, S);
[/asy]

The vector pointing from $(3,-10,1)$ to $(-5 + 4t, -3 - 18t, 6 - 4t)$ is then
\[\mathbf{v} = \begin{pmatrix} -8 + 4t \\ 7 - 18t \\ 5 - 4t \end{pmatrix}.\]For the point on the second line that is closest to $(3,-10,1),$ this vector will be orthogonal to the direction vector of the second line, which is $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}.$  Thus,
\[\begin{pmatrix} -8 + 4t \\ 7 - 18t \\ 5 - 4t \end{pmatrix} \cdot \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix} = 0.\]This gives us $(-8 + 4t)(4) + (7 - 18t)(-18) + (5 - 4t)(-4) = 0.$  Solving, we find $t = \frac{1}{2}.$

Substituting this value into $\mathbf{v},$ we find that the distance between the parallel lines is then
\[\|\mathbf{v}\| = \left\| \begin{pmatrix} -6 \\ -2 \\ 3 \end{pmatrix} \right\| = \boxed{7}.\]


Explanation:
Okay, so I need to find all values of p that satisfy both inequalities: 0 ≥ 54p - 144 and 0 > 12 - 20p. Hmm, let me start by solving each inequality separately and then find their intersection.

First, let me tackle the first inequality: 0 ≥ 54p - 144. I think I can rearrange this to solve for p. Let me subtract 54p from both sides? Wait, no, maybe I should add 144 to both sides first. Let me try that.

Starting with 0 ≥ 54p - 144. If I add 144 to both sides, that would give me 144 ≥ 54p. Then, to solve for p, I can divide both sides by 54. Let me do that.

So, 144 divided by 54... Hmm, let me simplify that fraction. Both numbers are divisible by 18? 144 divided by 18 is 8, and 54 divided by 18 is 3. So that simplifies to 8/3. Wait, so 144/54 is 8/3? Let me check: 54 times 8 is 432, and 144 times 3 is 432. Yes, so 144/54 reduces to 8/3. So, 8/3 ≥ p. Which is the same as p ≤ 8/3. Okay, so the first inequality simplifies to p ≤ 8/3. Got that.

Now, moving on to the second inequality: 0 > 12 - 20p. Let me solve for p here. Let me rearrange the inequality. First, I can subtract 12 from both sides? Wait, maybe it's easier to add 20p to both sides and then subtract 0? Let me think.

Starting with 0 > 12 - 20p. Let me add 20p to both sides. That would give me 20p > 12. Then, divide both sides by 20. So p > 12/20. Simplifying 12/20, divide numerator and denominator by 4: 3/5. So p > 3/5.

Wait, let me check that again. If I have 0 > 12 - 20p, then adding 20p to both sides gives 20p > 12. Yes, that's correct. Then dividing both sides by 20 gives p > 12/20, which simplifies to p > 3/5. Okay, so the second inequality is p > 3/5.

So now, I need to find all p that satisfy both p ≤ 8/3 and p > 3/5. So, in other words, p is greater than 3/5 and less than or equal to 8/3.

Let me write that as an interval. Since 3/5 is 0.6 and 8/3 is approximately 2.666..., the interval would be from 3/5 to 8/3, including 8/3 but not including 3/5. Wait, but hold on: the first inequality is p ≤ 8/3, which includes 8/3, and the second inequality is p > 3/5, which does not include 3/5. So the intersection would be all real numbers p such that 3/5 < p ≤ 8/3.

But let me confirm that I didn't make a mistake in solving the inequalities. Let me check the first inequality again. Original inequality: 0 ≥ 54p - 144. So, 54p - 144 ≤ 0. Adding 144 to both sides: 54p ≤ 144. Dividing by 54: p ≤ 144/54 = 8/3. Yes, that's correct.

Second inequality: 0 > 12 - 20p. Which is equivalent to 12 - 20p < 0. Subtract 12: -20p < -12. Then divide by -20, remembering to reverse the inequality sign when dividing by a negative number. So p > (-12)/(-20) = 12/20 = 3/5. Yes, that's correct. So the second inequality is p > 3/5.

Therefore, combining both inequalities, p must be greater than 3/5 and less than or equal to 8/3.

So in interval notation, that would be (3/5, 8/3].

But let me check if there are any fractions that need to be reduced. The problem says to reduce any fractions in the answer. 3/5 is already reduced, and 8/3 is also reduced. So the interval is (3/5, 8/3].

Wait, but let me make sure that there isn't a mistake in the direction of the inequalities. For example, when I solved the second inequality, I had 0 > 12 - 20p. Let me plug in a value to check. Let's take p = 1, which is greater than 3/5 (which is 0.6). Plugging into 12 - 20p: 12 - 20(1) = 12 - 20 = -8. Then 0 > -8 is true. So that works.

What if p = 3/5? Let's check. p = 3/5. Then 12 - 20*(3/5) = 12 - 12 = 0. So 0 > 0 is false. Therefore, p cannot be equal to 3/5, which matches our interval (3/5, 8/3].

Now check p = 8/3. Let's plug into the first inequality: 54*(8/3) - 144. Let me compute that. 54 divided by 3 is 18, so 18*8 = 144. Then 144 - 144 = 0. So 0 ≥ 0 is true. So p = 8/3 is included, which is correct.

What about a value between 3/5 and 8/3? Let's take p = 1. Let me check both inequalities. First inequality: 54*1 - 144 = 54 - 144 = -90. 0 ≥ -90 is true. Second inequality: 12 - 20*1 = -8, so 0 > -8 is true. So p = 1 is in the solution set.

What about p = 3/5 + a tiny bit? Let's say p = 0.61. Let me check the second inequality: 12 - 20*(0.61) = 12 - 12.2 = -0.2. So 0 > -0.2 is true. First inequality: 54*(0.61) - 144. Let me compute 54*0.6 = 32.4, 54*0.01 = 0.54, so total is 32.4 + 0.54 = 32.94. Then 32.94 - 144 = -111.06. 0 ≥ -111.06 is true. So that works.

What about p = 8/3? As before, that works. What about p = 3? Let me check. First inequality: 54*3 - 144 = 162 - 144 = 18. 0 ≥ 18? No, that's false. So p = 3 is not in the solution set, which is correct because 3 is greater than 8/3 (which is approx 2.666...).

Another check: p = 2. Let me check. First inequality: 54*2 - 144 = 108 - 144 = -36. 0 ≥ -36 is true. Second inequality: 12 - 20*2 = 12 - 40 = -28. 0 > -28 is true. So p = 2 is in the solution set.

So seems like the interval (3/5, 8/3] is correct.

But let me just make sure that there isn't a mistake in the direction of the inequality when solving. For example, when solving 0 > 12 - 20p, I added 20p to both sides and got 20p > 12. Alternatively, I could have subtracted 12 from both sides: 0 - 12 > -20p => -12 > -20p. Then divide both sides by -20, which reverses the inequality: (-12)/(-20) < p => 12/20 < p => 3/5 < p. Which is the same result. So that's correct.

So, yes, the solution set is all real numbers p such that 3/5 < p ≤ 8/3.

Therefore, in interval notation, that is (3/5, 8/3].

But let me check if the problem wants the answer in a specific format. The problem says "Express your answer in interval notation, reducing any fractions in your answer." So 3/5 and 8/3 are already reduced, so that's fine.

So I think that's the answer.

**Final Answer**
The solution in interval notation is \boxed{\left( \dfrac{3}{5}, \dfrac{8}{3} \right]}.

To solve the problem, we are given two inequalities:

1. $ 0 \ge 54p - 144 $
2. $ 0 > 12 - 20p $

We will solve each inequality separately and then find their **intersection**, since we are looking for values of $ p $ that satisfy **both** inequalities.

---

### Step 1: Solve the first inequality

$$
0 \ge 54p - 144
$$

Add 144 to both sides:

$$
144 \ge 54p
$$

Divide both sides by 54:

$$
\frac{144}{54} \ge p
$$

Simplify the fraction:

$$
\frac{144}{54} = \frac{8}{3}
$$

So the first inequality simplifies to:

$$
p \le \frac{8}{3}
$$

---

### Step 2: Solve the second inequality

$$
0 > 12 - 20p
$$

Subtract 12 from both sides:

$$
-12 > -20p
$$

Divide both sides by -20, **reversing the inequality**:

$$
\frac{-12}{-20} < p
$$

Simplify the fraction:

$$
\frac{12}{20} = \frac{3}{5}
$$

So the second inequality simplifies to:

$$
p > \frac{3}{5}
$$

---

### Step 3: Find the intersection of the two inequalities

We now have:

- $ p \le \frac{8}{3} $
- $ p > \frac{3}{5} $

The values of $ p $ that satisfy **both** are those that are **greater than** $ \frac{3}{5} $ and **less than or equal to** $ \frac{8}{3} $.

So the solution set is:

$$
\left( \frac{3}{5}, \frac{8}{3} \right]
$$

---

### Final Answer

$$
\boxed{\left( \dfrac{3}{5}, \dfrac{8}{3} \right]}
$$

[Assistant]
Explanation:
Okay, so I need to find all values of p that satisfy both inequalities: 0 ≥ 54p - 144 and 0 > 12 - 20p. Hmm, let me start by solving each inequality separately and then find their intersection.

First, let me tackle the first inequality: 0 ≥ 54p - 144. I think I can rearrange this to solve for p. Let me subtract 54p from both sides? Wait, no, maybe I should add 144 to both sides first. Let me try that.

Starting with 0 ≥ 54p - 144. If I add 144 to both sides, that would give me 144 ≥ 54p. Then, to solve for p, I can divide both sides by 54. Let me do that.

So, 144 divided by 54... Hmm, let me simplify that fraction. Both numbers are divisible by 18? 144 divided by 18 is 8, and 54 divided by 18 is 3. So that simplifies to 8/3. Wait, so 144/54 is 8/3? Let me check: 54 times 8 is 432, and 144 times 3 is 432. Yes, so 144/54 reduces to 8/3. So, 8/3 ≥ p. Which is the same as p ≤ 8/3. Okay, so the first inequality simplifies to p ≤ 8/3. Got that.

Now, moving on to the second inequality: 0 > 12 - 20p. Let me solve for p here. Let me rearrange the inequality. First, I can subtract 12 from both sides? Wait, maybe it's easier to add 20p to both sides and then subtract 0? Let me think.

Starting with 0 > 12 - 20p. Let me add 20p to both sides. That would give me 20p > 12. Then, divide both sides by 20. So p > 12/20. Simplifying 12/20, divide numerator and denominator by 4: 3/5. So p > 3/5.

Wait, let me check that again. If I have 0 > 12 - 20p, then adding 20p to both sides gives 20p > 12. Yes, that's correct. Then dividing both sides by 20 gives p > 12/20, which simplifies to p > 3/5. Okay, so the second inequality is p > 3/5.

So now, I need to find all p that satisfy both p ≤ 8/3 and p > 3/5. So, in other words, p is greater than 3/5 and less than or equal to 8/3.

Let me write that as an interval. Since 3/5 is 0.6 and 8/3 is approximately 2.666..., the interval would be from 3/5 to 8/3, including 8/3 but not including 3/5. Wait, but hold on: the first inequality is p ≤ 8/3, which includes 8/3, and the second inequality is p > 3/5, which does not include 3/5. So the intersection would be all real numbers p such that 3/5 < p ≤ 8/3.

But let me confirm that I didn't make a mistake in solving the inequalities. Let me check the first inequality again. Original inequality: 0 ≥ 54p - 144. So, 54p - 144 ≤ 0. Adding 144 to both sides: 54p ≤ 144. Dividing by 54: p ≤ 144/54 = 8/3. Yes, that's correct.

Second inequality: 0 > 12 - 20p. Which is equivalent to 12 - 20p < 0. Subtract 12: -20p < -12. Then divide by -20, remembering to reverse the inequality sign when dividing by a negative number. So p > (-12)/(-20) = 12/20 = 3/5. Yes, that's correct. So the second inequality is p > 3/5.

Therefore, combining both inequalities, p must be greater than 3/5 and less than or equal to 8/3.

So in interval notation, that would be (3/5, 8/3].

But let me check if there are any fractions that need to be reduced. The problem says to reduce any fractions in the answer. 3/5 is already reduced, and 8/3 is also reduced. So the interval is (3/5, 8/3].

Wait, but let me make sure that there isn't a mistake in the direction of the inequalities. For example, when I solved the second inequality, I had 0 > 12 - 20p. Let me plug in a value to check. Let's take p = 1, which is greater than 3/5 (which is 0.6). Plugging into 12 - 20p: 12 - 20(1) = 12 - 20 = -8. Then 0 > -8 is true. So that works.

What if p = 3/5? Let's check. p = 3/5. Then 12 - 20*(3/5) = 12 - 12 = 0. So 0 > 0 is false. Therefore, p cannot be equal to 3/5, which matches our interval (3/5, 8/3].

Now check p = 8/3. Let's plug into the first inequality: 54*(8/3) - 144. Let me compute that. 54 divided by 3 is 18, so 18*8 = 144. Then 144 - 144 = 0. So 0 ≥ 0 is true. So p = 8/3 is included, which is correct.

What about a value between 3/5 and 8/3? Let's take p = 1. Let me check both inequalities. First inequality: 54*1 - 144 = 54 - 144 = -90. 0 ≥ -90 is true. Second inequality: 12 - 20*1 = -8, so 0 > -8 is true. So p = 1 is in the solution set.

What about p = 3/5 + a tiny bit? Let's say p = 0.61. Let me check the second inequality: 12 - 20*(0.61) = 12 - 12.2 = -0.2. So 0 > -0.2 is true. First inequality: 54*(0.61) - 144. Let me compute 54*0.6 = 32.4, 54*0.01 = 0.54, so total is 32.4 + 0.54 = 32.94. Then 32.94 - 144 = -111.06. 0 ≥ -111.06 is true. So that works.

What about p = 8/3? As before, that works. What about p = 3? Let me check. First inequality: 54*3 - 144 = 162 - 144 = 18. 0 ≥ 18? No, that's false. So p = 3 is not in the solution set, which is correct because 3 is greater than 8/3 (which is approx 2.666...).

Another check: p = 2. Let me check. First inequality: 54*2 - 144 = 108 - 144 = -36. 0 ≥ -36 is true. Second inequality: 12 - 20*2 = 12 - 40 = -28. 0 > -28 is true. So p = 2 is in the solution set.

So seems like the interval (3/5, 8/3] is correct.

But let me just make sure that there isn't a mistake in the direction of the inequality when solving. For example, when solving 0 > 12 - 20p, I added 20p to both sides and got 20p > 12. Alternatively, I could have subtracted 12 from both sides: 0 - 12 > -20p => -12 > -20p. Then divide both sides by -20, which reverses the inequality: (-12)/(-20) < p => 12/20 < p => 3/5 < p. Which is the same result. So that's correct.

So, yes, the solution set is all real numbers p such that 3/5 < p ≤ 8/3.

Therefore, in interval notation, that is (3/5, 8/3].

But let me check if the problem wants the answer in a specific format. The problem says "Express your answer in interval notation, reducing any fractions in your answer." So 3/5 and 8/3 are already reduced, so that's fine.

So I think that's the answer.

**Final Answer**
The solution in interval notation is \boxed{\left( \dfrac{3}{5}, \dfrac{8}{3} \right]}.

To solve the problem, we are given two inequalities:

1. $ 0 \ge 54p - 144 $
2. $ 0 > 12 - 20p $

We will solve each inequality separately and then find their **intersection**, since we are looking for values of $ p $ that satisfy **both** inequalities.

---

### Step 1: Solve the first inequality

$$
0 \ge 54p - 144
$$

Add 144 to both sides:

$$
144 \ge 54p
$$

Divide both sides by 54:

$$
\frac{144}{54} \ge p
$$

Simplify the fraction:

$$
\frac{144}{54} = \frac{8}{3}
$$

So the first inequality simplifies to:

$$
p \le \frac{8}{3}
$$

---

### Step 2: Solve the second inequality

$$
0 > 12 - 20p
$$

Subtract 12 from both sides:

$$
-12 > -20p
$$

Divide both sides by -20, **reversing the inequality**:

$$
\frac{-12}{-20} < p
$$

Simplify the fraction:

$$
\frac{12}{20} = \frac{3}{5}
$$

So the second inequality simplifies to:

$$
p > \frac{3}{5}
$$

---

### Step 3: Find the intersection of the two inequalities

We now have:

- $ p \le \frac{8}{3} $
- $ p > \frac{3}{5} $

The values of $ p $ that satisfy **both** are those that are **greater than** $ \frac{3}{5} $ and **less than or equal to** $ \frac{8}{3} $.

So the solution set is:

$$
\left( \frac{3}{5}, \frac{8}{3} \right]
$$

---

### Final Answer

$$
\boxed{\left( \dfrac{3}{5}, \dfrac{8}{3} \right]}
$$

Solution:
We take the inequalities one at a time. Adding $144$ to both sides of the first inequality, we get $$144\ge 54p,$$implying $$\frac{144}{54}\ge p.$$Reducing the fraction and switching the sides (along with the direction of the inequality), we get $p\le\frac{8}{3}$.

To solve the second inequality, we add $20p$ to both sides: $$20p > 12$$Dividing both sides by $20$, we get $$p>\frac{12}{20}.$$Reducing the fraction gives $p>\frac{3}{5}$.

We are looking for $p$ which satisfy both inequalities. The intersection of the solutions above is $\boxed{\left(\frac{3}{5},\frac{8}{3}\right]}$.