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 next time when both marks are pointing north again. 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 each gear rotating independently, and we need to find when both marks are pointing north again?

Wait, but if they are separate gears, then their rotation periods are independent. So maybe the problem is similar to finding the least common multiple (LCM) of their rotation periods? Because the marks will point north again when each gear has completed an integer number of rotations. So, if I can find the time when both gears have completed an integer number of rotations, that would be when both marks are pointing north again.

But let me think again. Let me check the problem statement again: "One gear turns 33 1/3 times in a minute. Another gear turns 45 times in a minute. Initially, a mark on each gear is pointing due north. After how many seconds will the two gears next have both their marks pointing due north?"

So, the problem is about two gears, each rotating at different rates. The question is about when both marks will point north again. Since the gears are separate, their rotations are independent. Therefore, the time when both marks point north again is the least common multiple of their individual periods. So, first, I need to find the period of each gear, then find the LCM of those periods, and then convert that time into seconds.

But let me confirm. Let me think about the period. The period is the time it takes for one full rotation. So, for the first gear, which turns 33 1/3 times per minute, the period would be 1 minute divided by 33 1/3. Similarly for the second gear, which turns 45 times per minute, the period is 1 minute divided by 45.

Wait, but maybe I need to think in terms of rotations per second? Or maybe convert the rotation rates to rotations per second, then find the time when both have completed integer numbers of rotations?

Alternatively, since the marks point north again when each gear has completed an integer number of rotations. So, for the first gear, the time when the mark points north again is when the number of rotations is integer. Similarly for the second gear. Therefore, the time when both marks point north again is the least common multiple of their individual periods.

But let me think again. Let me take an example. Suppose gear A makes 1 rotation per minute, so its period is 1 minute. Gear B makes 2 rotations per minute, so its period is 0.5 minutes. Then, the LCM of 1 minute and 0.5 minutes is 1 minute. So, after 1 minute, both marks will point north again. Which makes sense, because gear A has done 1 rotation, gear B has done 2 rotations. So, both marks are back to the starting position. So, that seems correct.

Therefore, the same logic applies here. So, first, find the period of each gear, then find LCM of those periods, then convert that time into seconds.

So, let me compute the periods.

First gear: 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. Therefore, the period of the first gear is 1 / (100/3) minutes per rotation, which is 3/100 minutes per rotation. Similarly, the second gear is 45 rotations per minute, so its period is 1/45 minutes per rotation.

So, the periods are 3/100 minutes and 1/45 minutes. Now, I need to find the LCM of these two periods. But LCM of fractions... Hmm, how do I compute LCM of two fractions?

Wait, I remember that LCM of two numbers can be calculated using the formula:

LCM(a, b) = |a * b| / GCD(a, b)

But this is for integers. However, when dealing with fractions, there's a different approach. Wait, maybe I need to convert the periods into seconds first? Let me think.

Alternatively, since the LCM of two periods is the smallest time that is an integer multiple of both periods. So, if I can express the periods in terms of seconds, then find LCM of those times.

Let me try that. Let me convert both periods to seconds.

First gear: 3/100 minutes per rotation. Since 1 minute is 60 seconds, so 3/100 minutes is (3/100)*60 seconds. Let me compute that:

(3/100)*60 = (3*60)/100 = 180/100 = 9/5 seconds per rotation. Wait, that seems a bit strange. Wait, if the gear makes 100/3 rotations per minute, then the time per rotation is 1 / (100/3) minutes, which is 3/100 minutes. Converting that to seconds: 3/100 * 60 = 180/100 = 1.8 seconds per rotation. So, 1.8 seconds per rotation. Similarly, the second gear: 1/45 minutes per rotation. Converting to seconds: 1/45 * 60 = 60/45 = 4/3 seconds per rotation. So, approximately 1.333... seconds per rotation.

So, the periods are 1.8 seconds and 1.333... seconds. Now, I need to find the LCM of 1.8 and 1.333... seconds. But LCM of decimal numbers? Maybe it's easier to convert them to fractions.

So, 1.8 seconds is 9/5 seconds, and 4/3 seconds. So, LCM of 9/5 and 4/3. How do I compute LCM of fractions?

I recall that LCM of two fractions is given by LCM(numerators)/GCD(denominators). Wait, no, let me check.

Wait, there's a formula for LCM of fractions: LCM(a/b, c/d) = LCM(a, c) / GCD(b, d). Wait, is that correct? Wait, actually, I think the formula is LCM of the numerators divided by GCD of the denominators. Let me check with an example. Suppose I have 1/2 and 1/3. LCM of 1/2 and 1/3. The LCM should be 1, since after 1 minute, both would have completed integer rotations. Wait, but if I use the formula LCM(1,1)/GCD(2,3) = 1/1 = 1. Which is correct. So, maybe that formula is correct.

Alternatively, another way to think about LCM of fractions is to convert them to have the same denominator, then take LCM of the numerators and divide by the common denominator. Let me try that.

So, for 9/5 and 4/3. Let me find a common denominator. The denominators are 5 and 3, so LCM of 5 and 3 is 15. So, convert 9/5 to 27/15 and 4/3 to 20/15. Then, LCM of 27 and 20 is... Well, 27 and 20 are co-prime? 27 is 3^3, 20 is 2^2*5. So, LCM is 27*20 = 540. Therefore, LCM of 27/15 and 20/15 is 540/15 = 36. Therefore, LCM of 9/5 and 4/3 is 36 seconds? Wait, so that would mean that after 36 seconds, both gears will have completed integer number of rotations?

Wait, let me check. Let me compute how many rotations each gear makes in 36 seconds.

First gear: period is 9/5 seconds per rotation. So, number of rotations in 36 seconds is 36 / (9/5) = 36 * (5/9) = 4 * 5 = 20 rotations. Which is an integer. Good.

Second gear: period is 4/3 seconds per rotation. Number of rotations in 36 seconds is 36 / (4/3) = 36 * (3/4) = 9 * 3 = 27 rotations. Also an integer. Perfect. So, 36 seconds is the LCM. Therefore, the answer is 36 seconds?

Wait, but let me check if there is a smaller time. Let me think. Suppose there is a smaller time t where t is a multiple of both 9/5 and 4/3. So, t must satisfy t = k*(9/5) and t = m*(4/3), where k and m are integers. Therefore, k*(9/5) = m*(4/3). Therefore, 27k = 20m. Therefore, 27k = 20m. Therefore, the smallest integers k and m that satisfy this are k=20 and m=27. Therefore, t = 20*(9/5) = 36 seconds. So, yes, 36 seconds is the minimal time. Therefore, the answer is 36 seconds. Therefore, the answer is \boxed{36}?

Wait, but let me check again. Let me think in terms of rotations per second. Maybe another approach.

First gear: 33 1/3 rotations per minute. Let me convert that to rotations per second. 33 1/3 rotations per minute is equal to (100/3) rotations per minute, which is (100/3)/60 = 100/(3*60) = 100/180 = 5/9 rotations per second. Similarly, the second gear is 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 gear makes 3/4 rotations per second. Now, we need to find the time t when both gears have completed an integer number of rotations. So, for the first gear: number of rotations = (5/9)*t must be integer. For the second gear: number of rotations = (3/4)*t must be integer. Therefore, t must be such that (5/9)*t is integer and (3/4)*t is integer. Therefore, t must be a multiple of 9/5 and 4/3. Which is the same as before. Therefore, LCM of 9/5 and 4/3 is 36 seconds. Therefore, same answer.

Alternatively, think of t as the time when both gears have made integer numbers of rotations. So, t must satisfy:

(5/9)*t = integer => t = (9/5)*integer

(3/4)*t = integer => t = (4/3)*integer

Therefore, t must be a common multiple of 9/5 and 4/3. Therefore, the least such t is LCM(9/5, 4/3) = 36 seconds. So, same result.

Alternatively, think of the problem as finding t such that t is a multiple of both periods. Which is the same as above.

Alternatively, think in terms of angular speed. But maybe that complicates.

Alternatively, think of the problem as finding the time when both marks are pointing north again. Since the marks start pointing north, they will point north again after each gear has completed an integer number of rotations. Therefore, the time t must satisfy:

For gear 1: number of rotations = (33 1/3) * t / 60 = (100/3) * t / 60 = (100t)/180 = (5t)/9. Wait, no. Wait, if the gear makes 100/3 rotations per minute, then in t minutes, it makes (100/3)*t rotations. But we need the number of rotations to be integer. Wait, but if t is in minutes, then:

Wait, maybe I need to think in terms of t seconds. Let me try again.

Let me define t as the time in seconds. Then, the number of rotations for gear 1 is (number of rotations per minute) * (t / 60). So, gear 1: (100/3) * (t / 60) = (100t)/(180) = (5t)/9. Similarly, gear 2: 45 * (t / 60) = (45t)/60 = (3t)/4. Therefore, we need both (5t)/9 and (3t)/4 to be integers. Therefore, 5t/9 is integer and 3t/4 is integer. Therefore, t must be such that 5t is divisible by 9 and 3t is divisible by 4. Therefore:

From 5t divisible by 9: 5t ≡ 0 mod 9 => t ≡ 0 mod (9 / gcd(5,9)) => since gcd(5,9)=1, so t ≡ 0 mod 9. Therefore, t must be a multiple of 9.

From 3t divisible by 4: 3t ≡ 0 mod 4 => t ≡ 0 mod (4 / gcd(3,4)) => since gcd(3,4)=1, so t ≡ 0 mod 4. Therefore, t must be a multiple of 4.

Therefore, t must be a common multiple of 9 and 4. The least common multiple of 9 and 4 is 36. Therefore, t=36 seconds. Therefore, the answer is 36 seconds. So, same result again. Therefore, this seems to confirm that 36 seconds is the answer.

But let me check with actual numbers. Let me compute for t=36 seconds:

Gear 1: (5*36)/9 = 20 rotations. Which is integer.

Gear 2: (3*36)/4 = 27 rotations. Which is integer. Perfect. So, both marks point north again. And since 36 is the LCM of 9 and 4, there is no smaller t that satisfies both conditions. For example, check t= 36 / k, where k is integer greater than 1. Let's check t= 18 seconds:

Gear 1: (5*18)/9 = 10 rotations. Integer. Gear 2: (3*18)/4 = 54/4 = 13.5 rotations. Not integer. So, no good.

t= 12 seconds:

Gear 1: (5*12)/9 = 60/9 = 6.666... Not integer.

t= 9 seconds:

Gear 1: (5*9)/9 = 5 rotations. Integer. Gear 2: (3*9)/4 = 27/4 = 6.75. Not integer.

t= 4 seconds:

Gear 1: (5*4)/9 ≈ 2.222... Not integer.

t= 36 seconds is the first time when both are integers. Therefore, 36 seconds is correct.

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

But let me think again if there is another approach. For example, using the concept of relative speed or something else. Wait, but since the gears are not connected, their rotations are independent. Therefore, the problem reduces to finding the LCM of their individual periods, which we did. So, that seems correct.

Alternatively, think of the time when both marks point north again as the time when both gears have completed an integer number of rotations. Therefore, the problem is equivalent to solving for t such that:

Number of rotations of gear 1: (33 1/3) * t / 60 = integer.

Number of rotations of gear 2: 45 * t / 60 = integer.

Which is the same as before. Therefore, solving for t.

Alternatively, convert the rotation rates to revolutions per second:

Gear 1: 33 1/3 rev/min = 100/3 rev/min = 100/3 / 60 rev/sec = 100/180 = 5/9 rev/sec.

Gear 2: 45 rev/min = 45/60 = 3/4 rev/sec.

So, the time when both marks point north again is the least t such that 5/9 * t and 3/4 * t are integers. Which again leads us to t must be multiple of 9/5 and 4/3, so LCM(9/5, 4/3) = 36 seconds. So, same answer.

Alternatively, think of the problem as finding the LCM of the time periods. The time period for gear 1 is 1 / (5/9) = 9/5 seconds per revolution. The time period for gear 2 is 1 / (3/4) = 4/3 seconds per revolution. Therefore, LCM(9/5, 4/3) = 36 seconds. So, same result.

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

But just to make sure, let me check with another method. Let me think in terms of fractions.

We need t such that t is a multiple of both 9/5 and 4/3. So, t = 9/5 * k = 4/3 * m, where k and m are integers. Therefore, 9/5 * k = 4/3 * m => 27k = 20m. Therefore, 27k = 20m. Therefore, the smallest integers k and m that satisfy this equation are k=20 and m=27. Therefore, t = 9/5 * 20 = 36 seconds. Therefore, same answer. So, this is another way to think about it.

Therefore, I think I can be confident that the answer is 36 seconds.

**Final Answer**
\boxed{36}

To determine when both gears will have their marks pointing due north again, we need to analyze the rotational behavior of each gear and find the **least common multiple (LCM)** of their **rotation periods**.

---

### Step 1: Convert Rotational Speeds to Periods

We are given:

- **Gear A** rotates at $ 33\frac{1}{3} $ rotations per minute.
- **Gear B** rotates at 45 rotations per minute.

Convert these to **rotations per second**:

- Gear A:
  $$
  33\frac{1}{3} = \frac{100}{3} \text{ rotations per minute} = \frac{100}{3} \div 60 = \frac{5}{9} \text{ rotations per second}
  $$
  So, the **period** (time per rotation) is:
  $$
  \frac{1}{\frac{5}{9}} = \frac{9}{5} \text{ seconds per rotation}
  $$

- Gear B:
  $$
  45 \text{ rotations per minute} = \frac{45}{60} = \frac{3}{4} \text{ rotations per second}
  $$
  So, the **period** is:
  $$
  \frac{1}{\frac{3}{4}} = \frac{4}{3} \text{ seconds per rotation}
  $$

---

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

We now need to find the **least common multiple** of the two periods:

- Period of Gear A: $ \frac{9}{5} $ seconds
- Period of Gear B: $ \frac{4}{3} $ seconds

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 $ \frac{9}{5} $ and $ \frac{4}{3} $:

- LCM of numerators: $ \text{LCM}(9, 4) = 36 $
- GCD of denominators: $ \text{GCD}(5, 3) = 1 $

So,
$$
\text{LCM}\left(\frac{9}{5}, \frac{4}{3}\right) = \frac{36}{1} = 36 \text{ seconds}
$$

---

### Step 3: Verify the Result

Let’s confirm that after 36 seconds, both gears have completed an integer number of rotations:

- Gear A:
  $$
  \frac{36}{\frac{9}{5}} = 36 \times \frac{5}{9} = 20 \text{ full rotations}
  $$

- Gear B:
  $$
  \frac{36}{\frac{4}{3}} = 36 \times \frac{3}{4} = 27 \text{ full rotations}
  $$

Both are integers, so the marks will point due north again after **36 seconds**.

---

### 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 next time when both marks are pointing north again. 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 each gear rotating independently, and we need to find when both marks are pointing north again?

Wait, but if they are separate gears, then their rotation periods are independent. So maybe the problem is similar to finding the least common multiple (LCM) of their rotation periods? Because the marks will point north again when each gear has completed an integer number of rotations. So, if I can find the time when both gears have completed an integer number of rotations, that would be when both marks are pointing north again.

But let me think again. Let me check the problem statement again: "One gear turns 33 1/3 times in a minute. Another gear turns 45 times in a minute. Initially, a mark on each gear is pointing due north. After how many seconds will the two gears next have both their marks pointing due north?"

So, the problem is about two gears, each rotating at different rates. The question is about when both marks will point north again. Since the gears are separate, their rotations are independent. Therefore, the time when both marks point north again is the least common multiple of their individual periods. So, first, I need to find the period of each gear, then find the LCM of those periods, and then convert that time into seconds.

But let me confirm. Let me think about the period. The period is the time it takes for one full rotation. So, for the first gear, which turns 33 1/3 times per minute, the period would be 1 minute divided by 33 1/3. Similarly for the second gear, which turns 45 times per minute, the period is 1 minute divided by 45.

Wait, but maybe I need to think in terms of rotations per second? Or maybe convert the rotation rates to rotations per second, then find the time when both have completed integer numbers of rotations?

Alternatively, since the marks point north again when each gear has completed an integer number of rotations. So, for the first gear, the time when the mark points north again is when the number of rotations is integer. Similarly for the second gear. Therefore, the time when both marks point north again is the least common multiple of their individual periods.

But let me think again. Let me take an example. Suppose gear A makes 1 rotation per minute, so its period is 1 minute. Gear B makes 2 rotations per minute, so its period is 0.5 minutes. Then, the LCM of 1 minute and 0.5 minutes is 1 minute. So, after 1 minute, both marks will point north again. Which makes sense, because gear A has done 1 rotation, gear B has done 2 rotations. So, both marks are back to the starting position. So, that seems correct.

Therefore, the same logic applies here. So, first, find the period of each gear, then find LCM of those periods, then convert that time into seconds.

So, let me compute the periods.

First gear: 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. Therefore, the period of the first gear is 1 / (100/3) minutes per rotation, which is 3/100 minutes per rotation. Similarly, the second gear is 45 rotations per minute, so its period is 1/45 minutes per rotation.

So, the periods are 3/100 minutes and 1/45 minutes. Now, I need to find the LCM of these two periods. But LCM of fractions... Hmm, how do I compute LCM of two fractions?

Wait, I remember that LCM of two numbers can be calculated using the formula:

LCM(a, b) = |a * b| / GCD(a, b)

But this is for integers. However, when dealing with fractions, there's a different approach. Wait, maybe I need to convert the periods into seconds first? Let me think.

Alternatively, since the LCM of two periods is the smallest time that is an integer multiple of both periods. So, if I can express the periods in terms of seconds, then find LCM of those times.

Let me try that. Let me convert both periods to seconds.

First gear: 3/100 minutes per rotation. Since 1 minute is 60 seconds, so 3/100 minutes is (3/100)*60 seconds. Let me compute that:

(3/100)*60 = (3*60)/100 = 180/100 = 9/5 seconds per rotation. Wait, that seems a bit strange. Wait, if the gear makes 100/3 rotations per minute, then the time per rotation is 1 / (100/3) minutes, which is 3/100 minutes. Converting that to seconds: 3/100 * 60 = 180/100 = 1.8 seconds per rotation. So, 1.8 seconds per rotation. Similarly, the second gear: 1/45 minutes per rotation. Converting to seconds: 1/45 * 60 = 60/45 = 4/3 seconds per rotation. So, approximately 1.333... seconds per rotation.

So, the periods are 1.8 seconds and 1.333... seconds. Now, I need to find the LCM of 1.8 and 1.333... seconds. But LCM of decimal numbers? Maybe it's easier to convert them to fractions.

So, 1.8 seconds is 9/5 seconds, and 4/3 seconds. So, LCM of 9/5 and 4/3. How do I compute LCM of fractions?

I recall that LCM of two fractions is given by LCM(numerators)/GCD(denominators). Wait, no, let me check.

Wait, there's a formula for LCM of fractions: LCM(a/b, c/d) = LCM(a, c) / GCD(b, d). Wait, is that correct? Wait, actually, I think the formula is LCM of the numerators divided by GCD of the denominators. Let me check with an example. Suppose I have 1/2 and 1/3. LCM of 1/2 and 1/3. The LCM should be 1, since after 1 minute, both would have completed integer rotations. Wait, but if I use the formula LCM(1,1)/GCD(2,3) = 1/1 = 1. Which is correct. So, maybe that formula is correct.

Alternatively, another way to think about LCM of fractions is to convert them to have the same denominator, then take LCM of the numerators and divide by the common denominator. Let me try that.

So, for 9/5 and 4/3. Let me find a common denominator. The denominators are 5 and 3, so LCM of 5 and 3 is 15. So, convert 9/5 to 27/15 and 4/3 to 20/15. Then, LCM of 27 and 20 is... Well, 27 and 20 are co-prime? 27 is 3^3, 20 is 2^2*5. So, LCM is 27*20 = 540. Therefore, LCM of 27/15 and 20/15 is 540/15 = 36. Therefore, LCM of 9/5 and 4/3 is 36 seconds? Wait, so that would mean that after 36 seconds, both gears will have completed integer number of rotations?

Wait, let me check. Let me compute how many rotations each gear makes in 36 seconds.

First gear: period is 9/5 seconds per rotation. So, number of rotations in 36 seconds is 36 / (9/5) = 36 * (5/9) = 4 * 5 = 20 rotations. Which is an integer. Good.

Second gear: period is 4/3 seconds per rotation. Number of rotations in 36 seconds is 36 / (4/3) = 36 * (3/4) = 9 * 3 = 27 rotations. Also an integer. Perfect. So, 36 seconds is the LCM. Therefore, the answer is 36 seconds?

Wait, but let me check if there is a smaller time. Let me think. Suppose there is a smaller time t where t is a multiple of both 9/5 and 4/3. So, t must satisfy t = k*(9/5) and t = m*(4/3), where k and m are integers. Therefore, k*(9/5) = m*(4/3). Therefore, 27k = 20m. Therefore, 27k = 20m. Therefore, the smallest integers k and m that satisfy this are k=20 and m=27. Therefore, t = 20*(9/5) = 36 seconds. So, yes, 36 seconds is the minimal time. Therefore, the answer is 36 seconds. Therefore, the answer is \boxed{36}?

Wait, but let me check again. Let me think in terms of rotations per second. Maybe another approach.

First gear: 33 1/3 rotations per minute. Let me convert that to rotations per second. 33 1/3 rotations per minute is equal to (100/3) rotations per minute, which is (100/3)/60 = 100/(3*60) = 100/180 = 5/9 rotations per second. Similarly, the second gear is 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 gear makes 3/4 rotations per second. Now, we need to find the time t when both gears have completed an integer number of rotations. So, for the first gear: number of rotations = (5/9)*t must be integer. For the second gear: number of rotations = (3/4)*t must be integer. Therefore, t must be such that (5/9)*t is integer and (3/4)*t is integer. Therefore, t must be a multiple of 9/5 and 4/3. Which is the same as before. Therefore, LCM of 9/5 and 4/3 is 36 seconds. Therefore, same answer.

Alternatively, think of t as the time when both gears have made integer numbers of rotations. So, t must satisfy:

(5/9)*t = integer => t = (9/5)*integer

(3/4)*t = integer => t = (4/3)*integer

Therefore, t must be a common multiple of 9/5 and 4/3. Therefore, the least such t is LCM(9/5, 4/3) = 36 seconds. So, same result.

Alternatively, think of the problem as finding t such that t is a multiple of both periods. Which is the same as above.

Alternatively, think in terms of angular speed. But maybe that complicates.

Alternatively, think of the problem as finding the time when both marks are pointing north again. Since the marks start pointing north, they will point north again after each gear has completed an integer number of rotations. Therefore, the time t must satisfy:

For gear 1: number of rotations = (33 1/3) * t / 60 = (100/3) * t / 60 = (100t)/180 = (5t)/9. Wait, no. Wait, if the gear makes 100/3 rotations per minute, then in t minutes, it makes (100/3)*t rotations. But we need the number of rotations to be integer. Wait, but if t is in minutes, then:

Wait, maybe I need to think in terms of t seconds. Let me try again.

Let me define t as the time in seconds. Then, the number of rotations for gear 1 is (number of rotations per minute) * (t / 60). So, gear 1: (100/3) * (t / 60) = (100t)/(180) = (5t)/9. Similarly, gear 2: 45 * (t / 60) = (45t)/60 = (3t)/4. Therefore, we need both (5t)/9 and (3t)/4 to be integers. Therefore, 5t/9 is integer and 3t/4 is integer. Therefore, t must be such that 5t is divisible by 9 and 3t is divisible by 4. Therefore:

From 5t divisible by 9: 5t ≡ 0 mod 9 => t ≡ 0 mod (9 / gcd(5,9)) => since gcd(5,9)=1, so t ≡ 0 mod 9. Therefore, t must be a multiple of 9.

From 3t divisible by 4: 3t ≡ 0 mod 4 => t ≡ 0 mod (4 / gcd(3,4)) => since gcd(3,4)=1, so t ≡ 0 mod 4. Therefore, t must be a multiple of 4.

Therefore, t must be a common multiple of 9 and 4. The least common multiple of 9 and 4 is 36. Therefore, t=36 seconds. Therefore, the answer is 36 seconds. So, same result again. Therefore, this seems to confirm that 36 seconds is the answer.

But let me check with actual numbers. Let me compute for t=36 seconds:

Gear 1: (5*36)/9 = 20 rotations. Which is integer.

Gear 2: (3*36)/4 = 27 rotations. Which is integer. Perfect. So, both marks point north again. And since 36 is the LCM of 9 and 4, there is no smaller t that satisfies both conditions. For example, check t= 36 / k, where k is integer greater than 1. Let's check t= 18 seconds:

Gear 1: (5*18)/9 = 10 rotations. Integer. Gear 2: (3*18)/4 = 54/4 = 13.5 rotations. Not integer. So, no good.

t= 12 seconds:

Gear 1: (5*12)/9 = 60/9 = 6.666... Not integer.

t= 9 seconds:

Gear 1: (5*9)/9 = 5 rotations. Integer. Gear 2: (3*9)/4 = 27/4 = 6.75. Not integer.

t= 4 seconds:

Gear 1: (5*4)/9 ≈ 2.222... Not integer.

t= 36 seconds is the first time when both are integers. Therefore, 36 seconds is correct.

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

But let me think again if there is another approach. For example, using the concept of relative speed or something else. Wait, but since the gears are not connected, their rotations are independent. Therefore, the problem reduces to finding the LCM of their individual periods, which we did. So, that seems correct.

Alternatively, think of the time when both marks point north again as the time when both gears have completed an integer number of rotations. Therefore, the problem is equivalent to solving for t such that:

Number of rotations of gear 1: (33 1/3) * t / 60 = integer.

Number of rotations of gear 2: 45 * t / 60 = integer.

Which is the same as before. Therefore, solving for t.

Alternatively, convert the rotation rates to revolutions per second:

Gear 1: 33 1/3 rev/min = 100/3 rev/min = 100/3 / 60 rev/sec = 100/180 = 5/9 rev/sec.

Gear 2: 45 rev/min = 45/60 = 3/4 rev/sec.

So, the time when both marks point north again is the least t such that 5/9 * t and 3/4 * t are integers. Which again leads us to t must be multiple of 9/5 and 4/3, so LCM(9/5, 4/3) = 36 seconds. So, same answer.

Alternatively, think of the problem as finding the LCM of the time periods. The time period for gear 1 is 1 / (5/9) = 9/5 seconds per revolution. The time period for gear 2 is 1 / (3/4) = 4/3 seconds per revolution. Therefore, LCM(9/5, 4/3) = 36 seconds. So, same result.

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

But just to make sure, let me check with another method. Let me think in terms of fractions.

We need t such that t is a multiple of both 9/5 and 4/3. So, t = 9/5 * k = 4/3 * m, where k and m are integers. Therefore, 9/5 * k = 4/3 * m => 27k = 20m. Therefore, 27k = 20m. Therefore, the smallest integers k and m that satisfy this equation are k=20 and m=27. Therefore, t = 9/5 * 20 = 36 seconds. Therefore, same answer. So, this is another way to think about it.

Therefore, I think I can be confident that the answer is 36 seconds.

**Final Answer**
\boxed{36}

To determine when both gears will have their marks pointing due north again, we need to analyze the rotational behavior of each gear and find the **least common multiple (LCM)** of their **rotation periods**.

---

### Step 1: Convert Rotational Speeds to Periods

We are given:

- **Gear A** rotates at $ 33\frac{1}{3} $ rotations per minute.
- **Gear B** rotates at 45 rotations per minute.

Convert these to **rotations per second**:

- Gear A:
  $$
  33\frac{1}{3} = \frac{100}{3} \text{ rotations per minute} = \frac{100}{3} \div 60 = \frac{5}{9} \text{ rotations per second}
  $$
  So, the **period** (time per rotation) is:
  $$
  \frac{1}{\frac{5}{9}} = \frac{9}{5} \text{ seconds per rotation}
  $$

- Gear B:
  $$
  45 \text{ rotations per minute} = \frac{45}{60} = \frac{3}{4} \text{ rotations per second}
  $$
  So, the **period** is:
  $$
  \frac{1}{\frac{3}{4}} = \frac{4}{3} \text{ seconds per rotation}
  $$

---

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

We now need to find the **least common multiple** of the two periods:

- Period of Gear A: $ \frac{9}{5} $ seconds
- Period of Gear B: $ \frac{4}{3} $ seconds

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 $ \frac{9}{5} $ and $ \frac{4}{3} $:

- LCM of numerators: $ \text{LCM}(9, 4) = 36 $
- GCD of denominators: $ \text{GCD}(5, 3) = 1 $

So,
$$
\text{LCM}\left(\frac{9}{5}, \frac{4}{3}\right) = \frac{36}{1} = 36 \text{ seconds}
$$

---

### Step 3: Verify the Result

Let’s confirm that after 36 seconds, both gears have completed an integer number of rotations:

- Gear A:
  $$
  \frac{36}{\frac{9}{5}} = 36 \times \frac{5}{9} = 20 \text{ full rotations}
  $$

- Gear B:
  $$
  \frac{36}{\frac{4}{3}} = 36 \times \frac{3}{4} = 27 \text{ full rotations}
  $$

Both are integers, so the marks will point due north again after **36 seconds**.

---

### 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 interval notation. Since the first inequality is p ≤ 8/3 and the second is p > 3/5, the overlap is all numbers greater than 3/5 and less than or equal to 8/3. So in interval notation, that is (3/5, 8/3].

Yes, that seems right. 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, and **reverse the inequality** (since we divide by a negative number):

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

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} $.

In interval notation, this 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 interval notation. Since the first inequality is p ≤ 8/3 and the second is p > 3/5, the overlap is all numbers greater than 3/5 and less than or equal to 8/3. So in interval notation, that is (3/5, 8/3].

Yes, that seems right. 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, and **reverse the inequality** (since we divide by a negative number):

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

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} $.

In interval notation, this 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 $\boxed{\left(\frac{3}{5},\frac{8}{3}\right]}$.