Explanation:
Okay, let me try to figure out this gear problem. So, there are two gears. One turns 33 1/3 times per minute, and the other turns 45 times per minute. Initially, a mark on each gear is pointing due north. The question is asking: after how many seconds will both marks next point due north again?

Hmm. Alright, so I need to find the time when both gears have completed an integer number of rotations, right? Because if each gear has made a whole number of rotations, then their marks will be back to the starting position, which is pointing north. So, I need to find the least common multiple (LCM) of their rotation periods, and then convert that into seconds. But first, maybe I should convert their rotation rates into periods, which is the time per rotation. Then find the LCM of those periods?

Let me think. Let's start by converting the rotation rates to rotations per second, maybe? Or maybe to seconds per rotation?

Let me check: if the first gear turns 33 1/3 times in a minute, then its period (time per rotation) is 1 minute divided by 33 1/3 rotations. Similarly for the second gear. Then, the LCM of these two periods will give me the time when both gears are back to their starting position. Then convert that time into seconds?

Alternatively, maybe think in terms of how many rotations each gear makes in a certain time, and find the least common multiple of the number of rotations? Wait, that might not be straightforward. Let me think.

Let me recall that for gears, the number of rotations is related to their teeth, but here the problem doesn't mention teeth or anything like that. So maybe we need to just think about the rotation rates.

Let me try the first approach. Let me convert their rotation rates into rotations per minute, then find the time when both have completed an integer number of rotations.

Wait, but how does that work? If I have two gears rotating at different rates, how do I find when their marks both point north again?

Let me think. Suppose the first gear makes 33 1/3 rotations per minute. That's 100/3 rotations per minute, right? Because 33 1/3 is 100/3. So, the first gear's period T1 is 1 minute divided by 100/3 rotations, which is 3/100 minutes per rotation. Similarly, the second gear makes 45 rotations per minute, so its period T2 is 1/45 minutes per rotation.

Alternatively, maybe it's easier to convert these periods into seconds, so that the answer is in seconds. Let me try that.

So, first, converting T1 and T2 into seconds. Let's compute T1: 3/100 minutes per rotation. Convert minutes to seconds: 3/100 * 60 seconds = (3*60)/100 = 180/100 = 1.8 seconds per rotation. Similarly, T2 is 1/45 minutes per rotation. Convert to seconds: 1/45 * 60 = 60/45 = 4/3 ≈ 1.333... seconds per rotation.

So, T1 = 1.8 seconds per rotation, T2 = 4/3 seconds per rotation.

Now, we need to find the least common multiple of these two periods, so that both gears have completed an integer number of rotations. The LCM of T1 and T2 would give the time when both marks point north again.

But how do we compute LCM of two numbers that are decimal numbers? Maybe it's easier to convert them into fractions?

Let me think. T1 is 1.8 seconds, which is equal to 9/5 seconds. Because 1.8 = 9/5. Let me check: 9 divided by 5 is 1.8. Yes. And T2 is 4/3 seconds. So, we need to find LCM of 9/5 and 4/3.

Wait, how do you find LCM of fractions? Hmm. I think there is a formula for that. I recall that LCM of two fractions is given by LCM(numerators)/GCD(denominators). Wait, or is it something else?

Wait, actually, I think the formula is LCM(a/b, c/d) = LCM(a, c)/GCD(b, d). Wait, not sure. Maybe it's better to convert them into fractions and find the LCM.

Alternatively, think of LCM of the periods as the smallest time t such that t is an integer multiple of T1 and T2. So, t = k*T1 = m*T2, where k and m are integers. So, we need to find the smallest t such that t / T1 and t / T2 are integers.

Which would be equivalent to finding t such that t is a common multiple of T1 and T2, and the smallest such t. So, since T1 and T2 are in seconds, perhaps we can express them as fractions and then compute LCM?

Alternatively, since T1 is 9/5 seconds per rotation, and T2 is 4/3 seconds per rotation, then the number of rotations each gear makes in time t is t / T1 and t / T2. For both marks to be pointing north again, both t / T1 and t / T2 must be integers. So, t must be a multiple of T1 and T2. Therefore, t is LCM(T1, T2). But how do we compute LCM of fractions?

I think there's another way. Let me think. If we have two numbers, say a and b, then LCM(a, b) = (a*b)/GCD(a, b). But this is when they are integers. For fractions, maybe we can convert them to integers by multiplying by a common denominator?

Alternatively, think of LCM of T1 and T2 as the smallest t such that t is a multiple of both T1 and T2. So, t must be equal to k*T1 and m*T2. So, substituting, k*T1 = m*T2. Therefore, k / m = T2 / T1. Which is (4/3) / (9/5) = (4/3)*(5/9) = 20/27. So, k/m = 20/27. Therefore, the smallest integers k and m are 20 and 27. Therefore, t = 20*T1 = 20*(9/5) = 36 seconds, or t = 27*T2 = 27*(4/3) = 36 seconds. So, LCM is 36 seconds. Therefore, the answer is 36 seconds?

Wait, let me check again. Let me verify this.

First, T1 is 9/5 seconds per rotation. So, if t = 36 seconds, then the number of rotations for the first gear is 36 / (9/5) = 36 * (5/9) = 20 rotations. And for the second gear, t = 36 seconds, number of rotations is 36 / (4/3) = 36 * (3/4) = 27 rotations. So both are integers, which makes sense. Therefore, 36 seconds is the first time when both marks are pointing north again. Therefore, the answer is 36 seconds? But let me check if there's a smaller time.

Suppose, for example, that there's a smaller common multiple. Let's suppose that there is some t less than 36 seconds where both t / T1 and t / T2 are integers.

But since T1 = 9/5 and T2 = 4/3, let me think of t as a number of seconds. So, t must be a multiple of 9/5 and 4/3. Let me write t = 9/5 * k = 4/3 * m, where k and m are integers. Therefore, solving for t, we can find the least common multiple.

Alternatively, let me think of t as a multiple of both periods. So, if I consider the periods as fractions of seconds, then I need to find the least common multiple of 9/5 and 4/3.

There is a formula for LCM of two fractions: LCM(a/b, c/d) = LCM(a, c)/GCD(b, d). Wait, is that right? Wait, let me check with fractions. Wait, no, I think it's LCM(numerators)/GCD(denominators). Wait, but maybe not. Let me think of an example. For example, LCM(1/2, 1/3). What is that? The LCM would be the smallest number that is a multiple of 1/2 and 1/3. Which would be 1, since 1 is a multiple of both. But according to formula LCM(1,1)/GCD(2,3) = 1 / 1 = 1, which works. Another example: LCM(1/2, 1/4). The LCM should be 1/2. Let's check: formula would be LCM(1,1)/GCD(2,4)= 1 / 2 = 1/2. Which is correct. So maybe the formula is LCM of numerators over GCD of denominators. So, in general, LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). Let me check that.

So, in our case, the two periods are 9/5 and 4/3. So, numerators are 9 and 4. LCM(9,4) is 36. Denominators are 5 and 3. GCD(5,3) is 1. Therefore, LCM is 36 / 1 = 36. Therefore, the LCM is 36 seconds. Therefore, that's the answer. So, 36 seconds. Which matches with our previous calculation. Therefore, that seems correct.

Alternatively, another approach. Let me think about how many rotations each gear makes in t seconds. For the first gear, which makes 100/3 rotations per minute, that's (100/3) * (1/60) rotations per second. Wait, wait, let me check: 100/3 rotations per minute is equivalent to (100/3)/60 rotations per second. Which is 100/(3*60) = 100/180 = 5/9 rotations per second. Similarly, the second gear is 45 rotations per minute, so that's 45/60 = 3/4 rotations per second. Therefore, after t seconds, the first gear has made (5/9)*t rotations, and the second has made (3/4)*t rotations. For the marks to be pointing north again, both (5/9)*t and (3/4)*t must be integers. Therefore, t must be such that 5t/9 is integer and 3t/4 is integer. So, t must be a multiple of 9/5 and 4/3, as before. So, solving for t.

So, t must be such that t is a multiple of 9/5 and 4/3, which would mean t is LCM of 9/5 and 4/3. Which as before is 36.

Alternatively, since 5t/9 must be integer, so t must be a multiple of 9/5. Similarly, t must be a multiple of 4/3. Therefore, t must be the LCM of 9/5 and 4/3. Which we found as 36. So, that seems correct.

Alternatively, we can think in terms of when both gears have completed integer numbers of rotations. So, for gear 1: number of rotations is (100/3 rotations per minute) * (t/60 minutes) = (100/3)*(t/60) = (100t)/180 = (5t)/9. So, need 5t/9 integer. Similarly, for gear 2: 45 rotations per minute * (t/60) = (45t)/60 = (3t)/4. Need 3t/4 integer.

So, 5t/9 integer implies t must be multiple of 9/5. 3t/4 integer implies t must be multiple of 4/3. Therefore, t must be LCM(9/5, 4/3). Which is 36 seconds.

Alternatively, to solve equations 5t/9 = integer and 3t/4 = integer. Let me think about t. Let me let 5t/9 = a, and 3t/4 = b, where a and b are integers. Therefore, t = (9a)/5 and t = (4b)/3. Therefore, (9a)/5 = (4b)/3 => 27a = 20b. Therefore, 27a = 20b. So, the smallest integers a and b satisfying this are a = 20 and b = 27. Therefore, t = (9*20)/5 = 36, or t = (4*27)/3 = 36. So, same answer. So, t=36 seconds. Therefore, the answer is 36 seconds.

But wait, just to be thorough, let me check with actual numbers. Let me check when t = 36 seconds.

For the first gear: number of rotations = 5/9 * 36 = 20 rotations. Which is an integer. For the second gear: 3/4 * 36 = 27 rotations. Also integer. So, that works.

Is there a smaller t?

Let me suppose t is smaller. Let me suppose t is 18 seconds. Then, first gear: 5/9 * 18 = 10 rotations. That's integer. Second gear: 3/4 * 18 = 13.5 rotations. Not integer. So, not good.

t= 12 seconds. First gear: 5/9 *12 = 6.666... Not integer. So no.

t= 24 seconds. First gear: 5/9 *24 = 13.333... Not integer. Second gear: 3/4*24=18, which is integer. So, only one is integer.

t= 18 seconds, as above. First is 10, second is 13.5. Not good.

t= 36 seconds. Both integers. What about t= 9 seconds? First: 5/9 *9=5, integer. Second: 3/4*9=6.75, not integer.

t= 6 seconds: First: 5/9*6≈3.333, nope. Second: 3/4*6=4.5, nope.

t= 4 seconds: First: 5/9*4≈2.222, nope. Second: 3/4*4=3, integer.

t= 3 seconds: First: 5/9*3≈1.666, nope. Second: 3/4*3=2.25, nope.

t= 5 seconds: First: 5/9*5≈2.777, nope. Second: 3/4*5=3.75, nope.

t= 8 seconds: First: 5/9*8≈4.444, nope. Second: 3/4*8=6, integer.

So, only at t=36 seconds both are integers. So yes, 36 seconds is the smallest. Therefore, the answer is 36 seconds. So, that seems correct.

Alternatively, maybe think in terms of minutes. Let me try that approach. Maybe convert the problem into minutes.

First gear makes 100/3 rotations per minute. Second gear makes 45 rotations per minute. So, find the least time t (in minutes) such that both gears have completed integer number of rotations.

So, 100/3 * t = integer, and 45 * t = integer. Let me denote t as some time in minutes. So, for the first gear: t must be multiple of 3/100 minutes. Because 100/3 rotations per minute, so period is 3/100 minutes per rotation. Therefore, t must be multiple of 3/100. Similarly, the second gear has period 1/45 minutes per rotation, so t must be multiple of 1/45. Therefore, LCM(3/100, 1/45) ?

Again, same problem. Find LCM of 3/100 and 1/45. Which would be 3/100 * (1/45) / GCD(3/100, 1/45)? Wait, perhaps better to convert into fractions again. Let me think.

Alternatively, express t as LCM(3/100, 1/45). Let me use the formula for LCM of fractions. Which is LCM(numerators)/GCD(denominators). Wait, but numerators are 3 and 1? Wait, no, the fractions are 3/100 and 1/45. So, if we consider the formula LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). Wait, then LCM of 3 and 1 is 3. GCD of 100 and 45 is 5. So, LCM would be 3 / 5 = 3/5 minutes. Then, converting that to seconds: 3/5 minutes * 60 seconds/minute = 36 seconds. So, same answer. Therefore, that also gives 36 seconds. So, that confirms again. So, with that method, LCM of 3/100 and 1/45 is 3/5 minutes, which is 36 seconds. So, same result. Therefore, 36 seconds.

Therefore, I think 36 seconds is the correct answer. Let me check again with the initial approach.

So, gear 1 makes 33 1/3 rotations per minute. Which is 100/3 rotations per minute. So, in one minute, gear 1 does 100/3 rotations. Gear 2 does 45 rotations per minute. So, the question is, after how many seconds will both gears have made an integer number of rotations?

So, the problem reduces to find the smallest t such that (100/3)*t is integer and 45*t is integer. Wait, but t is in minutes? Or in seconds?

Wait, if we take t in minutes, then for gear 1: number of rotations is (100/3)*t. For gear 2: 45*t. So, need (100/3)*t and 45*t to be integers. So, let me find t in minutes.

Let me set t such that:

(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. Which we already found is 3/5 minutes. Which is 36 seconds. So, same answer. Therefore, this approach also gives 36 seconds.

Alternatively, suppose we wanted to find t in seconds. Let me think. Let me convert the rotation rates to rotations per second. Gear 1: 100/3 rotations per minute is equal to (100/3)/60 = 100/180 = 5/9 rotations per second. Gear 2: 45 rotations per minute is 45/60 = 3/4 rotations per second. Then, the number of rotations after t seconds is (5/9)*t and (3/4)*t. We need both to be integers. Therefore, same as before, so t must satisfy 5t/9 and 3t/4 integers. Which gives t = 36 seconds. So, same answer. Therefore, all approaches point to 36 seconds. Therefore, I think that is the answer.

But just to make sure, let me think of another way. Let me think about the time it takes for each gear to complete a full rotation, then find the least common multiple of those times. Which is the approach I took earlier.

First gear: 1 rotation takes 3/100 minutes, which is 1.8 seconds. Second gear: 1 rotation takes 1/45 minutes, which is 4/3 seconds ≈ 1.333 seconds. So, LCM of 1.8 seconds and 1.333 seconds. Let me convert these to fractions. 1.8 seconds is 9/5 seconds, 4/3 seconds. LCM of 9/5 and 4/3. As before, LCM(9/5, 4/3) = LCM(9,4)/GCD(5,3) = 36 / 1 = 36. Therefore, 36 seconds. Same result. So, all approaches confirm that.

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 next have their marks pointing due north, we analyze the rotational behavior of both gears over time.

---

### **Step 1: Interpret the Given Rates**

- Gear 1 turns **33 1/3 rotations per minute**
  Converting to an improper fraction:
  $$
  33\frac{1}{3} = \frac{100}{3} \text{ rotations per minute}
  $$

- Gear 2 turns **45 rotations per minute**

---

### **Step 2: Convert Rotation Rates to Periods (Time per Rotation)**

We want to determine how long it takes for each gear to complete **one full rotation**, i.e., the **period** of each gear.

- **Gear 1**:
  $$
  \text{Period}_1 = \frac{1}{\text{Rotations per minute}} = \frac{1}{\frac{100}{3}} = \frac{3}{100} \text{ minutes}
  $$
  Convert to seconds:
  $$
  \frac{3}{100} \times 60 = \frac{180}{100} = \frac{9}{5} = 1.8 \text{ seconds per rotation}
  $$

- **Gear 2**:
  $$
  \text{Period}_2 = \frac{1}{45} \text{ minutes} = \frac{1}{45} \times 60 = \frac{4}{3} \approx 1.333 \text{ seconds per rotation}
  $$

---

### **Step 3: Find the Least Common Multiple (LCM) of the Periods**

We now seek the **least common multiple** of the two periods, which will give the first time both gears simultaneously return to their initial position (i.e., the mark pointing north again).

- The periods in seconds are:
  - $ \text{Period}_1 = \frac{9}{5} $
  - $ \text{Period}_2 = \frac{4}{3} $

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 formula:
- Numerators: $ a = 9 $, $ c = 4 $, so $ \text{LCM}(9,4) = 36 $
- Denominators: $ b = 5 $, $ d = 3 $, so $ \text{GCD}(5,3) = 1 $
$$
\text{LCM}\left(\frac{9}{5}, \frac{4}{3}\right) = \frac{36}{1} = 36 \text{ seconds}
$$

---

### **Step 4: Verification**

- In **36 seconds**, Gear 1 makes:
  $$
  \frac{36}{\frac{9}{5}} = 36 \times \frac{5}{9} = 20 \text{ full rotations}
  $$

- In **36 seconds**, Gear 2 makes:
  $$
  \frac{36}{\frac{4}{3}} = 36 \times \frac{3}{4} = 27 \text{ full rotations}
  $$

Since both have completed an integer number of rotations, the marks will both be back to point north again after **36 seconds**.

---

### **Final Answer**
$$
\boxed{36}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this gear problem. So, there are two gears. One turns 33 1/3 times per minute, and the other turns 45 times per minute. Initially, a mark on each gear is pointing due north. The question is asking: after how many seconds will both marks next point due north again?

Hmm. Alright, so I need to find the time when both gears have completed an integer number of rotations, right? Because if each gear has made a whole number of rotations, then their marks will be back to the starting position, which is pointing north. So, I need to find the least common multiple (LCM) of their rotation periods, and then convert that into seconds. But first, maybe I should convert their rotation rates into periods, which is the time per rotation. Then find the LCM of those periods?

Let me think. Let's start by converting the rotation rates to rotations per second, maybe? Or maybe to seconds per rotation?

Let me check: if the first gear turns 33 1/3 times in a minute, then its period (time per rotation) is 1 minute divided by 33 1/3 rotations. Similarly for the second gear. Then, the LCM of these two periods will give me the time when both gears are back to their starting position. Then convert that time into seconds?

Alternatively, maybe think in terms of how many rotations each gear makes in a certain time, and find the least common multiple of the number of rotations? Wait, that might not be straightforward. Let me think.

Let me recall that for gears, the number of rotations is related to their teeth, but here the problem doesn't mention teeth or anything like that. So maybe we need to just think about the rotation rates.

Let me try the first approach. Let me convert their rotation rates into rotations per minute, then find the time when both have completed an integer number of rotations.

Wait, but how does that work? If I have two gears rotating at different rates, how do I find when their marks both point north again?

Let me think. Suppose the first gear makes 33 1/3 rotations per minute. That's 100/3 rotations per minute, right? Because 33 1/3 is 100/3. So, the first gear's period T1 is 1 minute divided by 100/3 rotations, which is 3/100 minutes per rotation. Similarly, the second gear makes 45 rotations per minute, so its period T2 is 1/45 minutes per rotation.

Alternatively, maybe it's easier to convert these periods into seconds, so that the answer is in seconds. Let me try that.

So, first, converting T1 and T2 into seconds. Let's compute T1: 3/100 minutes per rotation. Convert minutes to seconds: 3/100 * 60 seconds = (3*60)/100 = 180/100 = 1.8 seconds per rotation. Similarly, T2 is 1/45 minutes per rotation. Convert to seconds: 1/45 * 60 = 60/45 = 4/3 ≈ 1.333... seconds per rotation.

So, T1 = 1.8 seconds per rotation, T2 = 4/3 seconds per rotation.

Now, we need to find the least common multiple of these two periods, so that both gears have completed an integer number of rotations. The LCM of T1 and T2 would give the time when both marks point north again.

But how do we compute LCM of two numbers that are decimal numbers? Maybe it's easier to convert them into fractions?

Let me think. T1 is 1.8 seconds, which is equal to 9/5 seconds. Because 1.8 = 9/5. Let me check: 9 divided by 5 is 1.8. Yes. And T2 is 4/3 seconds. So, we need to find LCM of 9/5 and 4/3.

Wait, how do you find LCM of fractions? Hmm. I think there is a formula for that. I recall that LCM of two fractions is given by LCM(numerators)/GCD(denominators). Wait, or is it something else?

Wait, actually, I think the formula is LCM(a/b, c/d) = LCM(a, c)/GCD(b, d). Wait, not sure. Maybe it's better to convert them into fractions and find the LCM.

Alternatively, think of LCM of the periods as the smallest time t such that t is an integer multiple of T1 and T2. So, t = k*T1 = m*T2, where k and m are integers. So, we need to find the smallest t such that t / T1 and t / T2 are integers.

Which would be equivalent to finding t such that t is a common multiple of T1 and T2, and the smallest such t. So, since T1 and T2 are in seconds, perhaps we can express them as fractions and then compute LCM?

Alternatively, since T1 is 9/5 seconds per rotation, and T2 is 4/3 seconds per rotation, then the number of rotations each gear makes in time t is t / T1 and t / T2. For both marks to be pointing north again, both t / T1 and t / T2 must be integers. So, t must be a multiple of T1 and T2. Therefore, t is LCM(T1, T2). But how do we compute LCM of fractions?

I think there's another way. Let me think. If we have two numbers, say a and b, then LCM(a, b) = (a*b)/GCD(a, b). But this is when they are integers. For fractions, maybe we can convert them to integers by multiplying by a common denominator?

Alternatively, think of LCM of T1 and T2 as the smallest t such that t is a multiple of both T1 and T2. So, t must be equal to k*T1 and m*T2. So, substituting, k*T1 = m*T2. Therefore, k / m = T2 / T1. Which is (4/3) / (9/5) = (4/3)*(5/9) = 20/27. So, k/m = 20/27. Therefore, the smallest integers k and m are 20 and 27. Therefore, t = 20*T1 = 20*(9/5) = 36 seconds, or t = 27*T2 = 27*(4/3) = 36 seconds. So, LCM is 36 seconds. Therefore, the answer is 36 seconds?

Wait, let me check again. Let me verify this.

First, T1 is 9/5 seconds per rotation. So, if t = 36 seconds, then the number of rotations for the first gear is 36 / (9/5) = 36 * (5/9) = 20 rotations. And for the second gear, t = 36 seconds, number of rotations is 36 / (4/3) = 36 * (3/4) = 27 rotations. So both are integers, which makes sense. Therefore, 36 seconds is the first time when both marks are pointing north again. Therefore, the answer is 36 seconds? But let me check if there's a smaller time.

Suppose, for example, that there's a smaller common multiple. Let's suppose that there is some t less than 36 seconds where both t / T1 and t / T2 are integers.

But since T1 = 9/5 and T2 = 4/3, let me think of t as a number of seconds. So, t must be a multiple of 9/5 and 4/3. Let me write t = 9/5 * k = 4/3 * m, where k and m are integers. Therefore, solving for t, we can find the least common multiple.

Alternatively, let me think of t as a multiple of both periods. So, if I consider the periods as fractions of seconds, then I need to find the least common multiple of 9/5 and 4/3.

There is a formula for LCM of two fractions: LCM(a/b, c/d) = LCM(a, c)/GCD(b, d). Wait, is that right? Wait, let me check with fractions. Wait, no, I think it's LCM(numerators)/GCD(denominators). Wait, but maybe not. Let me think of an example. For example, LCM(1/2, 1/3). What is that? The LCM would be the smallest number that is a multiple of 1/2 and 1/3. Which would be 1, since 1 is a multiple of both. But according to formula LCM(1,1)/GCD(2,3) = 1 / 1 = 1, which works. Another example: LCM(1/2, 1/4). The LCM should be 1/2. Let's check: formula would be LCM(1,1)/GCD(2,4)= 1 / 2 = 1/2. Which is correct. So maybe the formula is LCM of numerators over GCD of denominators. So, in general, LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). Let me check that.

So, in our case, the two periods are 9/5 and 4/3. So, numerators are 9 and 4. LCM(9,4) is 36. Denominators are 5 and 3. GCD(5,3) is 1. Therefore, LCM is 36 / 1 = 36. Therefore, the LCM is 36 seconds. Therefore, that's the answer. So, 36 seconds. Which matches with our previous calculation. Therefore, that seems correct.

Alternatively, another approach. Let me think about how many rotations each gear makes in t seconds. For the first gear, which makes 100/3 rotations per minute, that's (100/3) * (1/60) rotations per second. Wait, wait, let me check: 100/3 rotations per minute is equivalent to (100/3)/60 rotations per second. Which is 100/(3*60) = 100/180 = 5/9 rotations per second. Similarly, the second gear is 45 rotations per minute, so that's 45/60 = 3/4 rotations per second. Therefore, after t seconds, the first gear has made (5/9)*t rotations, and the second has made (3/4)*t rotations. For the marks to be pointing north again, both (5/9)*t and (3/4)*t must be integers. Therefore, t must be such that 5t/9 is integer and 3t/4 is integer. So, t must be a multiple of 9/5 and 4/3, as before. So, solving for t.

So, t must be such that t is a multiple of 9/5 and 4/3, which would mean t is LCM of 9/5 and 4/3. Which as before is 36.

Alternatively, since 5t/9 must be integer, so t must be a multiple of 9/5. Similarly, t must be a multiple of 4/3. Therefore, t must be the LCM of 9/5 and 4/3. Which we found as 36. So, that seems correct.

Alternatively, we can think in terms of when both gears have completed integer numbers of rotations. So, for gear 1: number of rotations is (100/3 rotations per minute) * (t/60 minutes) = (100/3)*(t/60) = (100t)/180 = (5t)/9. So, need 5t/9 integer. Similarly, for gear 2: 45 rotations per minute * (t/60) = (45t)/60 = (3t)/4. Need 3t/4 integer.

So, 5t/9 integer implies t must be multiple of 9/5. 3t/4 integer implies t must be multiple of 4/3. Therefore, t must be LCM(9/5, 4/3). Which is 36 seconds.

Alternatively, to solve equations 5t/9 = integer and 3t/4 = integer. Let me think about t. Let me let 5t/9 = a, and 3t/4 = b, where a and b are integers. Therefore, t = (9a)/5 and t = (4b)/3. Therefore, (9a)/5 = (4b)/3 => 27a = 20b. Therefore, 27a = 20b. So, the smallest integers a and b satisfying this are a = 20 and b = 27. Therefore, t = (9*20)/5 = 36, or t = (4*27)/3 = 36. So, same answer. So, t=36 seconds. Therefore, the answer is 36 seconds.

But wait, just to be thorough, let me check with actual numbers. Let me check when t = 36 seconds.

For the first gear: number of rotations = 5/9 * 36 = 20 rotations. Which is an integer. For the second gear: 3/4 * 36 = 27 rotations. Also integer. So, that works.

Is there a smaller t?

Let me suppose t is smaller. Let me suppose t is 18 seconds. Then, first gear: 5/9 * 18 = 10 rotations. That's integer. Second gear: 3/4 * 18 = 13.5 rotations. Not integer. So, not good.

t= 12 seconds. First gear: 5/9 *12 = 6.666... Not integer. So no.

t= 24 seconds. First gear: 5/9 *24 = 13.333... Not integer. Second gear: 3/4*24=18, which is integer. So, only one is integer.

t= 18 seconds, as above. First is 10, second is 13.5. Not good.

t= 36 seconds. Both integers. What about t= 9 seconds? First: 5/9 *9=5, integer. Second: 3/4*9=6.75, not integer.

t= 6 seconds: First: 5/9*6≈3.333, nope. Second: 3/4*6=4.5, nope.

t= 4 seconds: First: 5/9*4≈2.222, nope. Second: 3/4*4=3, integer.

t= 3 seconds: First: 5/9*3≈1.666, nope. Second: 3/4*3=2.25, nope.

t= 5 seconds: First: 5/9*5≈2.777, nope. Second: 3/4*5=3.75, nope.

t= 8 seconds: First: 5/9*8≈4.444, nope. Second: 3/4*8=6, integer.

So, only at t=36 seconds both are integers. So yes, 36 seconds is the smallest. Therefore, the answer is 36 seconds. So, that seems correct.

Alternatively, maybe think in terms of minutes. Let me try that approach. Maybe convert the problem into minutes.

First gear makes 100/3 rotations per minute. Second gear makes 45 rotations per minute. So, find the least time t (in minutes) such that both gears have completed integer number of rotations.

So, 100/3 * t = integer, and 45 * t = integer. Let me denote t as some time in minutes. So, for the first gear: t must be multiple of 3/100 minutes. Because 100/3 rotations per minute, so period is 3/100 minutes per rotation. Therefore, t must be multiple of 3/100. Similarly, the second gear has period 1/45 minutes per rotation, so t must be multiple of 1/45. Therefore, LCM(3/100, 1/45) ?

Again, same problem. Find LCM of 3/100 and 1/45. Which would be 3/100 * (1/45) / GCD(3/100, 1/45)? Wait, perhaps better to convert into fractions again. Let me think.

Alternatively, express t as LCM(3/100, 1/45). Let me use the formula for LCM of fractions. Which is LCM(numerators)/GCD(denominators). Wait, but numerators are 3 and 1? Wait, no, the fractions are 3/100 and 1/45. So, if we consider the formula LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). Wait, then LCM of 3 and 1 is 3. GCD of 100 and 45 is 5. So, LCM would be 3 / 5 = 3/5 minutes. Then, converting that to seconds: 3/5 minutes * 60 seconds/minute = 36 seconds. So, same answer. Therefore, that also gives 36 seconds. So, that confirms again. So, with that method, LCM of 3/100 and 1/45 is 3/5 minutes, which is 36 seconds. So, same result. Therefore, 36 seconds.

Therefore, I think 36 seconds is the correct answer. Let me check again with the initial approach.

So, gear 1 makes 33 1/3 rotations per minute. Which is 100/3 rotations per minute. So, in one minute, gear 1 does 100/3 rotations. Gear 2 does 45 rotations per minute. So, the question is, after how many seconds will both gears have made an integer number of rotations?

So, the problem reduces to find the smallest t such that (100/3)*t is integer and 45*t is integer. Wait, but t is in minutes? Or in seconds?

Wait, if we take t in minutes, then for gear 1: number of rotations is (100/3)*t. For gear 2: 45*t. So, need (100/3)*t and 45*t to be integers. So, let me find t in minutes.

Let me set t such that:

(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. Which we already found is 3/5 minutes. Which is 36 seconds. So, same answer. Therefore, this approach also gives 36 seconds.

Alternatively, suppose we wanted to find t in seconds. Let me think. Let me convert the rotation rates to rotations per second. Gear 1: 100/3 rotations per minute is equal to (100/3)/60 = 100/180 = 5/9 rotations per second. Gear 2: 45 rotations per minute is 45/60 = 3/4 rotations per second. Then, the number of rotations after t seconds is (5/9)*t and (3/4)*t. We need both to be integers. Therefore, same as before, so t must satisfy 5t/9 and 3t/4 integers. Which gives t = 36 seconds. So, same answer. Therefore, all approaches point to 36 seconds. Therefore, I think that is the answer.

But just to make sure, let me think of another way. Let me think about the time it takes for each gear to complete a full rotation, then find the least common multiple of those times. Which is the approach I took earlier.

First gear: 1 rotation takes 3/100 minutes, which is 1.8 seconds. Second gear: 1 rotation takes 1/45 minutes, which is 4/3 seconds ≈ 1.333 seconds. So, LCM of 1.8 seconds and 1.333 seconds. Let me convert these to fractions. 1.8 seconds is 9/5 seconds, 4/3 seconds. LCM of 9/5 and 4/3. As before, LCM(9/5, 4/3) = LCM(9,4)/GCD(5,3) = 36 / 1 = 36. Therefore, 36 seconds. Same result. So, all approaches confirm that.

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 next have their marks pointing due north, we analyze the rotational behavior of both gears over time.

---

### **Step 1: Interpret the Given Rates**

- Gear 1 turns **33 1/3 rotations per minute**
  Converting to an improper fraction:
  $$
  33\frac{1}{3} = \frac{100}{3} \text{ rotations per minute}
  $$

- Gear 2 turns **45 rotations per minute**

---

### **Step 2: Convert Rotation Rates to Periods (Time per Rotation)**

We want to determine how long it takes for each gear to complete **one full rotation**, i.e., the **period** of each gear.

- **Gear 1**:
  $$
  \text{Period}_1 = \frac{1}{\text{Rotations per minute}} = \frac{1}{\frac{100}{3}} = \frac{3}{100} \text{ minutes}
  $$
  Convert to seconds:
  $$
  \frac{3}{100} \times 60 = \frac{180}{100} = \frac{9}{5} = 1.8 \text{ seconds per rotation}
  $$

- **Gear 2**:
  $$
  \text{Period}_2 = \frac{1}{45} \text{ minutes} = \frac{1}{45} \times 60 = \frac{4}{3} \approx 1.333 \text{ seconds per rotation}
  $$

---

### **Step 3: Find the Least Common Multiple (LCM) of the Periods**

We now seek the **least common multiple** of the two periods, which will give the first time both gears simultaneously return to their initial position (i.e., the mark pointing north again).

- The periods in seconds are:
  - $ \text{Period}_1 = \frac{9}{5} $
  - $ \text{Period}_2 = \frac{4}{3} $

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 formula:
- Numerators: $ a = 9 $, $ c = 4 $, so $ \text{LCM}(9,4) = 36 $
- Denominators: $ b = 5 $, $ d = 3 $, so $ \text{GCD}(5,3) = 1 $
$$
\text{LCM}\left(\frac{9}{5}, \frac{4}{3}\right) = \frac{36}{1} = 36 \text{ seconds}
$$

---

### **Step 4: Verification**

- In **36 seconds**, Gear 1 makes:
  $$
  \frac{36}{\frac{9}{5}} = 36 \times \frac{5}{9} = 20 \text{ full rotations}
  $$

- In **36 seconds**, Gear 2 makes:
  $$
  \frac{36}{\frac{4}{3}} = 36 \times \frac{3}{4} = 27 \text{ full rotations}
  $$

Since both have completed an integer number of rotations, the marks will both be back to point 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 this. I remember that when two lines are parallel, the distance between them can be found by taking any point on one line and then finding the distance from that point to the other line. Since they are parallel, this distance should be the same no matter which point I choose on the first line.

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

The first line is given by the point $\begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}$ plus $t$ times the direction vector $\begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$.

The second line is $\begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}$ plus $u$ times the direction vector $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$.

Wait, the problem says they are parallel. Let me check if their direction vectors are scalar multiples of each other. For the first direction vector, it's $\begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$, and the second one is $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$. If I multiply the first direction vector by 2, I get $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$, which is exactly the second direction vector. So yes, they are scalar multiples (specifically, multiplied by 2), so the lines are indeed parallel. That checks out.

Now, to find the distance between two parallel lines. Let me recall the formula. If I have two lines, one with point $\mathbf{a}$ and direction vector $\mathbf{v}$, and the other with point $\mathbf{b}$ and the same direction vector $\mathbf{v}$, then the distance between them is the norm of the cross product of the vector $\mathbf{b} - \mathbf{a}$ and the direction vector $\mathbf{v}$, divided by the norm of $\mathbf{v}$.

So, formula-wise: distance = $\frac{|\mathbf{(b - a)} \times \mathbf{v}|}{|\mathbf{v}|}$.

Alternatively, I remember sometimes you can take a point from one line, find the vector connecting that point to the other line, and then project that vector onto a vector perpendicular to the direction of the lines. Since they are parallel, the direction vector is the same.

But maybe the cross product method is more straightforward. Let me try that.

So let me denote the first line as Line 1: $\mathbf{r}_1 = \mathbf{a} + t\mathbf{v}$, where $\mathbf{a} = \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}$, and $\mathbf{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$.

The second line is Line 2: $\mathbf{r}_2 = \mathbf{b} + u\mathbf{v}$, where $\mathbf{b} = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}$.

So, vector $\mathbf{b} - \mathbf{a}$ is $\begin{pmatrix} -5 - 3 \\ -3 - (-10) \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$.

Then, compute the cross product of $\mathbf{b} - \mathbf{a}$ and $\mathbf{v}$. Let me write that out.

Let me denote $\mathbf{w} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$, and $\mathbf{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$.

So, $\mathbf{w} \times \mathbf{v}$.

The cross product can be calculated using determinant:

$$
\mathbf{w} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-8 & 7 & 5 \\
2 & -9 & -2 \\
\end{vmatrix}
$$

Calculating this determinant:

First component (i): $7 \cdot (-2) - 5 \cdot (-9) = -14 + 45 = 31$

Second component (j): - [ (-8) \cdot (-2) - 5 \cdot 2 ] = - [16 - 10] = -6

Third component (k): (-8) \cdot (-9) - 7 \cdot 2 = 72 - 14 = 58

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

Now, the magnitude of this cross product is $\sqrt{31^2 + (-6)^2 + 58^2}$.

Calculating each term:

31 squared: 961

(-6) squared: 36

58 squared: 3364

Sum: 961 + 36 + 3364 = let's add them step by step. 961 + 36 is 997, then 997 + 3364 is 4361.

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

Wait, let me verify my cross product calculation again.

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

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

Third component: (-8)*(-9) - 7*2 = 72 - 14 = 58. Correct. So cross product vector is indeed (31, -6, 58).

So magnitude squared is 31² + (-6)² + 58². Let me confirm 31²:

30² is 900, 31² is 961. Correct.

58²: Well, 60² is 3600, so 58² is (60 - 2)² = 60² - 4*60 + 4 = 3600 - 240 + 4 = 3364. Correct.

So 961 + 36 is 997. 997 + 3364 is 4361. So sqrt(4361). Hmm. Let me check if 4361 is a perfect square.

What's 66²? 4356. 67² is 4489. So sqrt(4361) is between 66 and 67.

Wait, but maybe there's a calculation mistake? Because 66² is 4356, so 4361 - 4356 = 5. So sqrt(4361) = 66 + 5/(2*66) approximately, but exact value is irrational. Maybe I need to keep it as sqrt(4361) for now.

Then, the magnitude of direction vector v is sqrt(2² + (-9)² + (-2)²) = sqrt(4 + 81 + 4) = sqrt(89).

Therefore, the distance is sqrt(4361)/sqrt(89).

Hmm, maybe we can simplify sqrt(4361)/sqrt(89) as sqrt(4361/89). Let me compute 4361 divided by 89.

Let me check if 89 divides into 4361 evenly. Let me compute 89 * 49 = 89*(50 - 1) = 4450 - 89 = 4361. Oh! 89 * 49 = 4361.

Wow, that's a nice number. So sqrt(4361 / 89) = sqrt(49) = 7.

Wait, that's a big simplification! Let me check again:

If 89 * 49 = 4361, then 4361 divided by 89 is indeed 49. Therefore, sqrt(4361)/sqrt(89) = sqrt(49) = 7.

So the distance between the two lines is 7.

Wait, that seems straightforward. Let me verify once again to make sure I didn't make a mistake in cross product or in the calculation of the vector b - a.

Vector b - a: original points are (3, -10, 1) and (-5, -3, 6). So subtracting coordinates: (-5 - 3, -3 - (-10), 6 - 1) = (-8, 7, 5). Correct.

Cross product with v = (2, -9, -2) gives (31, -6, 58). Then magnitude squared is 31² + (-6)^2 + 58² = 961 + 36 + 3364 = 4361. Then, |v| is sqrt(89).

Then, sqrt(4361)/sqrt(89) = sqrt(4361/89) = sqrt(49) = 7.

Yes, that seems correct. Therefore, the distance between the two lines is 7.

But let me think again, maybe there is another way to check. For example, take a point on the first line and find the distance to the second line. Let me try that method to confirm.

Take a point from the first line, say when t = 0: point A is (3, -10, 1). Then, find the distance from point A to the second line.

The distance from a point to a line in 3D can be calculated using the formula: distance = |(vector from point to line) cross direction vector| / |direction vector|.

So, vector from point A to a point on the second line. Let me take point B on the second line, which is (-5, -3, 6). So vector AB is (-5 - 3, -3 - (-10), 6 - 1) = (-8, 7, 5), same as before. Then cross product with direction vector v is same as before. So same result. So that gives same distance.

Alternatively, maybe take another point? Let me take a different point on the first line. Let me say t = 1: (3 + 2, -10 - 9, 1 - 2) = (5, -19, -1). Let me compute the distance from this point to the second line.

Let me denote this point as P = (5, -19, -1). The second line is given by point B = (-5, -3, 6) and direction vector v = (2, -9, -2).

So vector from P to B is (-5 - 5, -3 - (-19), 6 - (-1)) = (-10, 16, 7). Then cross product with direction vector v:

Compute (-10, 16, 7) × (2, -9, -2). Let me compute:

i component: 16*(-2) - 7*(-9) = -32 + 63 = 31

j component: - [ (-10)*(-2) - 7*2 ] = - [20 - 14] = -6

k component: (-10)*(-9) - 16*2 = 90 - 32 = 58

So again cross product is (31, -6, 58), same as before. The magnitude is sqrt(4361), divide by sqrt(89) gives 7. So same result. So that confirms the answer.

Alternatively, maybe I can parametrize one line and find the minimum distance. But that might be more complicated, but let me try.

Let me parametrize the first line as (3 + 2t, -10 -9t, 1 - 2t). The second line is (-5 + 4u, -3 -18u, 6 -4u).

To find the distance between these two lines, we can find the minimum distance between any two points on the lines. So we need to minimize the distance squared between (3 + 2t, -10 -9t, 1 - 2t) and (-5 + 4u, -3 -18u, 6 -4u).

Let me denote the distance squared as:

D² = [ (3 + 2t - (-5 + 4u))² + (-10 -9t - (-3 -18u))² + (1 - 2t - (6 -4u))² ]

Simplify each term:

First coordinate difference: 3 + 2t +5 -4u = 8 + 2t -4u

Second coordinate difference: -10 -9t +3 +18u = -7 -9t +18u

Third coordinate difference: 1 -2t -6 +4u = -5 -2t +4u

So D² = (8 + 2t -4u)² + (-7 -9t +18u)² + (-5 -2t +4u)²

Now, since the lines are parallel, the minimal distance should be achieved when the vector connecting the points is perpendicular to the direction vector of the lines.

Alternatively, since the lines are parallel, the minimal distance occurs when the line segment connecting the two points is perpendicular to both lines. Therefore, the vector connecting the two points should be perpendicular to the direction vector of the lines.

So, if we denote vector between points as (8 + 2t -4u, -7 -9t +18u, -5 -2t +4u). This vector must be perpendicular to direction vector v = (2, -9, -2).

Therefore, their dot product must be zero:

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

Let me compute each term:

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

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

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

Adding all three terms:

16 + 4t -8u + 63 +81t -162u +10 +4t -8u

Combine like terms:

Constants: 16 + 63 + 10 = 89

t terms: 4t +81t +4t = 89t

u terms: -8u -162u -8u = -178u

So equation: 89 + 89t -178u = 0

Simplify: 89(t - 2u) + 89 = 0 => 89(t - 2u + 1) = 0 => t - 2u + 1 = 0 => t = 2u -1

So this is a constraint. Now, we can use this relation to substitute into the expression for D² and then minimize.

Let me substitute t = 2u -1 into D². Let me first write down D² again:

D² = (8 + 2t -4u)^2 + (-7 -9t +18u)^2 + (-5 -2t +4u)^2

Substituting t = 2u -1:

First term: 8 + 2*(2u -1) -4u = 8 +4u -2 -4u = 6

Second term: -7 -9*(2u -1) +18u = -7 -18u +9 +18u = (-7 +9) + (-18u +18u) = 2 + 0 = 2

Third term: -5 -2*(2u -1) +4u = -5 -4u +2 +4u = (-5 + 2) + (-4u +4u) = -3 + 0 = -3

So D² = (6)^2 + (2)^2 + (-3)^2 = 36 + 4 + 9 = 49. Therefore, D = sqrt(49) = 7.

Wow, so that gives me the same answer. Therefore, the minimal distance between the two lines is indeed 7.

So both methods confirm that the distance is 7. Therefore, the answer is 7.

But just to make sure, let me check with another approach. For instance, since the lines are parallel, we can translate one of the lines so that it passes through the origin, and then compute the distance.

But maybe that's more complicated. Alternatively, since we have the cross product method and the parametrization method both leading to 7, I think that's solid.

Alternatively, could there be an error in assuming the formula? Let me recall the formula for the distance between two parallel lines.

Yes, the formula is indeed |(b - a) × v| / |v|. So since I did that and also verified through parametrization, that's good.

Another way: since the direction vector is v = (2, -9, -2), then if I can find a vector connecting the two lines, which is (b - a) = (-8, 7, 5), and then project this vector onto a direction perpendicular to v. The length of that projection is the distance.

The projection length is |(b - a) × v| / |v|, which is exactly the formula I used. Therefore, that's consistent.

Therefore, I think the answer is definitely 7.

Just to recap, the steps were:

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

2. Use the cross product formula for the distance between two parallel lines, which involves taking the cross product of (b - a) and direction vector, then dividing by the magnitude of direction vector.

3. Calculated cross product, found magnitude squared, divided by magnitude of direction vector squared, simplified and got 7.

4. Verified using parametrization method by setting up equations for minimal distance, found that the minimal distance squared is 49, so distance is 7.

Therefore, confident that the answer is 7.

**Final Answer**
\boxed{7}

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

> **If two lines are parallel**, the shortest distance between them is the perpendicular distance from any point on one line to the other. This can be calculated using the formula:

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

Where:
- $\vec{w} = \vec{b} - \vec{a}$ is the vector between a point on the first line $\vec{a}$ and a point on the second line $\vec{b}$,
- $\vec{v}$ is the common direction vector of both lines.

---

### Step 1: Identify points and direction vectors

Given:
- First line:
  $\vec{a} = \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}$,
  $\vec{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$

- Second line:
  $\vec{b} = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}$,
  $\vec{v} = \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$ (which is $2 \cdot \vec{v}$, so the lines are **parallel**)

---

### Step 2: Compute $\vec{w} = \vec{b} - \vec{a}$

$$
\vec{w} = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}
$$

---

### Step 3: Compute the cross product $\vec{w} \times \vec{v}$

Let’s calculate:
$$
\vec{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}
\quad
\vec{w} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}
$$

$$
\vec{w} \times \vec{v} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-8 & 7 & 5 \\
2 & -9 & -2 \\
\end{vmatrix}
= \begin{pmatrix}
7 \cdot (-2) - 5 \cdot (-9) \\
-( -8 \cdot (-2) - 5 \cdot 2 ) \\
-8 \cdot (-9) - 7 \cdot 2
\end{pmatrix}
= \begin{pmatrix} -14 + 45 \\ -(16 - 10) \\ 72 - 14 \end{pmatrix}
= \begin{pmatrix} 31 \\ -6 \\ 58 \end{pmatrix}
$$

---

### Step 4: Compute the magnitudes

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

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

---

### Step 5: Compute the distance

$$
\text{Distance} = \frac{|\vec{w} \times \vec{v}|}{|\vec{v}|} = \frac{\sqrt{4361}}{\sqrt{89}}
$$

We observe that:

$$
\sqrt{4361} = \sqrt{89 \cdot 49} = \sqrt{89} \cdot \sqrt{49} = \sqrt{89} \cdot 7
$$

So,

$$
\frac{\sqrt{4361}}{\sqrt{89}} = \frac{\sqrt{89} \cdot 7}{\sqrt{89}} = 7
$$

---

### ✅ Final Answer

$$
\boxed{7}
$$

[Assistant]
Explanation:
Okay, so I need to find the distance between two parallel lines. Let me recall how to do this. I remember that when two lines are parallel, the distance between them can be found by taking any point on one line and then finding the distance from that point to the other line. Since they are parallel, this distance should be the same no matter which point I choose on the first line.

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

The first line is given by the point $\begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}$ plus $t$ times the direction vector $\begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$.

The second line is $\begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}$ plus $u$ times the direction vector $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$.

Wait, the problem says they are parallel. Let me check if their direction vectors are scalar multiples of each other. For the first direction vector, it's $\begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$, and the second one is $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$. If I multiply the first direction vector by 2, I get $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$, which is exactly the second direction vector. So yes, they are scalar multiples (specifically, multiplied by 2), so the lines are indeed parallel. That checks out.

Now, to find the distance between two parallel lines. Let me recall the formula. If I have two lines, one with point $\mathbf{a}$ and direction vector $\mathbf{v}$, and the other with point $\mathbf{b}$ and the same direction vector $\mathbf{v}$, then the distance between them is the norm of the cross product of the vector $\mathbf{b} - \mathbf{a}$ and the direction vector $\mathbf{v}$, divided by the norm of $\mathbf{v}$.

So, formula-wise: distance = $\frac{|\mathbf{(b - a)} \times \mathbf{v}|}{|\mathbf{v}|}$.

Alternatively, I remember sometimes you can take a point from one line, find the vector connecting that point to the other line, and then project that vector onto a vector perpendicular to the direction of the lines. Since they are parallel, the direction vector is the same.

But maybe the cross product method is more straightforward. Let me try that.

So let me denote the first line as Line 1: $\mathbf{r}_1 = \mathbf{a} + t\mathbf{v}$, where $\mathbf{a} = \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}$, and $\mathbf{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$.

The second line is Line 2: $\mathbf{r}_2 = \mathbf{b} + u\mathbf{v}$, where $\mathbf{b} = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}$.

So, vector $\mathbf{b} - \mathbf{a}$ is $\begin{pmatrix} -5 - 3 \\ -3 - (-10) \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$.

Then, compute the cross product of $\mathbf{b} - \mathbf{a}$ and $\mathbf{v}$. Let me write that out.

Let me denote $\mathbf{w} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}$, and $\mathbf{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$.

So, $\mathbf{w} \times \mathbf{v}$.

The cross product can be calculated using determinant:

$$
\mathbf{w} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-8 & 7 & 5 \\
2 & -9 & -2 \\
\end{vmatrix}
$$

Calculating this determinant:

First component (i): $7 \cdot (-2) - 5 \cdot (-9) = -14 + 45 = 31$

Second component (j): - [ (-8) \cdot (-2) - 5 \cdot 2 ] = - [16 - 10] = -6

Third component (k): (-8) \cdot (-9) - 7 \cdot 2 = 72 - 14 = 58

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

Now, the magnitude of this cross product is $\sqrt{31^2 + (-6)^2 + 58^2}$.

Calculating each term:

31 squared: 961

(-6) squared: 36

58 squared: 3364

Sum: 961 + 36 + 3364 = let's add them step by step. 961 + 36 is 997, then 997 + 3364 is 4361.

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

Wait, let me verify my cross product calculation again.

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

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

Third component: (-8)*(-9) - 7*2 = 72 - 14 = 58. Correct. So cross product vector is indeed (31, -6, 58).

So magnitude squared is 31² + (-6)² + 58². Let me confirm 31²:

30² is 900, 31² is 961. Correct.

58²: Well, 60² is 3600, so 58² is (60 - 2)² = 60² - 4*60 + 4 = 3600 - 240 + 4 = 3364. Correct.

So 961 + 36 is 997. 997 + 3364 is 4361. So sqrt(4361). Hmm. Let me check if 4361 is a perfect square.

What's 66²? 4356. 67² is 4489. So sqrt(4361) is between 66 and 67.

Wait, but maybe there's a calculation mistake? Because 66² is 4356, so 4361 - 4356 = 5. So sqrt(4361) = 66 + 5/(2*66) approximately, but exact value is irrational. Maybe I need to keep it as sqrt(4361) for now.

Then, the magnitude of direction vector v is sqrt(2² + (-9)² + (-2)²) = sqrt(4 + 81 + 4) = sqrt(89).

Therefore, the distance is sqrt(4361)/sqrt(89).

Hmm, maybe we can simplify sqrt(4361)/sqrt(89) as sqrt(4361/89). Let me compute 4361 divided by 89.

Let me check if 89 divides into 4361 evenly. Let me compute 89 * 49 = 89*(50 - 1) = 4450 - 89 = 4361. Oh! 89 * 49 = 4361.

Wow, that's a nice number. So sqrt(4361 / 89) = sqrt(49) = 7.

Wait, that's a big simplification! Let me check again:

If 89 * 49 = 4361, then 4361 divided by 89 is indeed 49. Therefore, sqrt(4361)/sqrt(89) = sqrt(49) = 7.

So the distance between the two lines is 7.

Wait, that seems straightforward. Let me verify once again to make sure I didn't make a mistake in cross product or in the calculation of the vector b - a.

Vector b - a: original points are (3, -10, 1) and (-5, -3, 6). So subtracting coordinates: (-5 - 3, -3 - (-10), 6 - 1) = (-8, 7, 5). Correct.

Cross product with v = (2, -9, -2) gives (31, -6, 58). Then magnitude squared is 31² + (-6)^2 + 58² = 961 + 36 + 3364 = 4361. Then, |v| is sqrt(89).

Then, sqrt(4361)/sqrt(89) = sqrt(4361/89) = sqrt(49) = 7.

Yes, that seems correct. Therefore, the distance between the two lines is 7.

But let me think again, maybe there is another way to check. For example, take a point on the first line and find the distance to the second line. Let me try that method to confirm.

Take a point from the first line, say when t = 0: point A is (3, -10, 1). Then, find the distance from point A to the second line.

The distance from a point to a line in 3D can be calculated using the formula: distance = |(vector from point to line) cross direction vector| / |direction vector|.

So, vector from point A to a point on the second line. Let me take point B on the second line, which is (-5, -3, 6). So vector AB is (-5 - 3, -3 - (-10), 6 - 1) = (-8, 7, 5), same as before. Then cross product with direction vector v is same as before. So same result. So that gives same distance.

Alternatively, maybe take another point? Let me take a different point on the first line. Let me say t = 1: (3 + 2, -10 - 9, 1 - 2) = (5, -19, -1). Let me compute the distance from this point to the second line.

Let me denote this point as P = (5, -19, -1). The second line is given by point B = (-5, -3, 6) and direction vector v = (2, -9, -2).

So vector from P to B is (-5 - 5, -3 - (-19), 6 - (-1)) = (-10, 16, 7). Then cross product with direction vector v:

Compute (-10, 16, 7) × (2, -9, -2). Let me compute:

i component: 16*(-2) - 7*(-9) = -32 + 63 = 31

j component: - [ (-10)*(-2) - 7*2 ] = - [20 - 14] = -6

k component: (-10)*(-9) - 16*2 = 90 - 32 = 58

So again cross product is (31, -6, 58), same as before. The magnitude is sqrt(4361), divide by sqrt(89) gives 7. So same result. So that confirms the answer.

Alternatively, maybe I can parametrize one line and find the minimum distance. But that might be more complicated, but let me try.

Let me parametrize the first line as (3 + 2t, -10 -9t, 1 - 2t). The second line is (-5 + 4u, -3 -18u, 6 -4u).

To find the distance between these two lines, we can find the minimum distance between any two points on the lines. So we need to minimize the distance squared between (3 + 2t, -10 -9t, 1 - 2t) and (-5 + 4u, -3 -18u, 6 -4u).

Let me denote the distance squared as:

D² = [ (3 + 2t - (-5 + 4u))² + (-10 -9t - (-3 -18u))² + (1 - 2t - (6 -4u))² ]

Simplify each term:

First coordinate difference: 3 + 2t +5 -4u = 8 + 2t -4u

Second coordinate difference: -10 -9t +3 +18u = -7 -9t +18u

Third coordinate difference: 1 -2t -6 +4u = -5 -2t +4u

So D² = (8 + 2t -4u)² + (-7 -9t +18u)² + (-5 -2t +4u)²

Now, since the lines are parallel, the minimal distance should be achieved when the vector connecting the points is perpendicular to the direction vector of the lines.

Alternatively, since the lines are parallel, the minimal distance occurs when the line segment connecting the two points is perpendicular to both lines. Therefore, the vector connecting the two points should be perpendicular to the direction vector of the lines.

So, if we denote vector between points as (8 + 2t -4u, -7 -9t +18u, -5 -2t +4u). This vector must be perpendicular to direction vector v = (2, -9, -2).

Therefore, their dot product must be zero:

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

Let me compute each term:

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

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

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

Adding all three terms:

16 + 4t -8u + 63 +81t -162u +10 +4t -8u

Combine like terms:

Constants: 16 + 63 + 10 = 89

t terms: 4t +81t +4t = 89t

u terms: -8u -162u -8u = -178u

So equation: 89 + 89t -178u = 0

Simplify: 89(t - 2u) + 89 = 0 => 89(t - 2u + 1) = 0 => t - 2u + 1 = 0 => t = 2u -1

So this is a constraint. Now, we can use this relation to substitute into the expression for D² and then minimize.

Let me substitute t = 2u -1 into D². Let me first write down D² again:

D² = (8 + 2t -4u)^2 + (-7 -9t +18u)^2 + (-5 -2t +4u)^2

Substituting t = 2u -1:

First term: 8 + 2*(2u -1) -4u = 8 +4u -2 -4u = 6

Second term: -7 -9*(2u -1) +18u = -7 -18u +9 +18u = (-7 +9) + (-18u +18u) = 2 + 0 = 2

Third term: -5 -2*(2u -1) +4u = -5 -4u +2 +4u = (-5 + 2) + (-4u +4u) = -3 + 0 = -3

So D² = (6)^2 + (2)^2 + (-3)^2 = 36 + 4 + 9 = 49. Therefore, D = sqrt(49) = 7.

Wow, so that gives me the same answer. Therefore, the minimal distance between the two lines is indeed 7.

So both methods confirm that the distance is 7. Therefore, the answer is 7.

But just to make sure, let me check with another approach. For instance, since the lines are parallel, we can translate one of the lines so that it passes through the origin, and then compute the distance.

But maybe that's more complicated. Alternatively, since we have the cross product method and the parametrization method both leading to 7, I think that's solid.

Alternatively, could there be an error in assuming the formula? Let me recall the formula for the distance between two parallel lines.

Yes, the formula is indeed |(b - a) × v| / |v|. So since I did that and also verified through parametrization, that's good.

Another way: since the direction vector is v = (2, -9, -2), then if I can find a vector connecting the two lines, which is (b - a) = (-8, 7, 5), and then project this vector onto a direction perpendicular to v. The length of that projection is the distance.

The projection length is |(b - a) × v| / |v|, which is exactly the formula I used. Therefore, that's consistent.

Therefore, I think the answer is definitely 7.

Just to recap, the steps were:

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

2. Use the cross product formula for the distance between two parallel lines, which involves taking the cross product of (b - a) and direction vector, then dividing by the magnitude of direction vector.

3. Calculated cross product, found magnitude squared, divided by magnitude of direction vector squared, simplified and got 7.

4. Verified using parametrization method by setting up equations for minimal distance, found that the minimal distance squared is 49, so distance is 7.

Therefore, confident that the answer is 7.

**Final Answer**
\boxed{7}

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

> **If two lines are parallel**, the shortest distance between them is the perpendicular distance from any point on one line to the other. This can be calculated using the formula:

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

Where:
- $\vec{w} = \vec{b} - \vec{a}$ is the vector between a point on the first line $\vec{a}$ and a point on the second line $\vec{b}$,
- $\vec{v}$ is the common direction vector of both lines.

---

### Step 1: Identify points and direction vectors

Given:
- First line:
  $\vec{a} = \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix}$,
  $\vec{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}$

- Second line:
  $\vec{b} = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix}$,
  $\vec{v} = \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}$ (which is $2 \cdot \vec{v}$, so the lines are **parallel**)

---

### Step 2: Compute $\vec{w} = \vec{b} - \vec{a}$

$$
\vec{w} = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}
$$

---

### Step 3: Compute the cross product $\vec{w} \times \vec{v}$

Let’s calculate:
$$
\vec{v} = \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}
\quad
\vec{w} = \begin{pmatrix} -8 \\ 7 \\ 5 \end{pmatrix}
$$

$$
\vec{w} \times \vec{v} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-8 & 7 & 5 \\
2 & -9 & -2 \\
\end{vmatrix}
= \begin{pmatrix}
7 \cdot (-2) - 5 \cdot (-9) \\
-( -8 \cdot (-2) - 5 \cdot 2 ) \\
-8 \cdot (-9) - 7 \cdot 2
\end{pmatrix}
= \begin{pmatrix} -14 + 45 \\ -(16 - 10) \\ 72 - 14 \end{pmatrix}
= \begin{pmatrix} 31 \\ -6 \\ 58 \end{pmatrix}
$$

---

### Step 4: Compute the magnitudes

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

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

---

### Step 5: Compute the distance

$$
\text{Distance} = \frac{|\vec{w} \times \vec{v}|}{|\vec{v}|} = \frac{\sqrt{4361}}{\sqrt{89}}
$$

We observe that:

$$
\sqrt{4361} = \sqrt{89 \cdot 49} = \sqrt{89} \cdot \sqrt{49} = \sqrt{89} \cdot 7
$$

So,

$$
\frac{\sqrt{4361}}{\sqrt{89}} = \frac{\sqrt{89} \cdot 7}{\sqrt{89}} = 7
$$

---

### ✅ 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. Let me write them down again to make sure I have them right.

First inequality: 0 ≥ 54p - 144. Second inequality: 0 > 12 - 20p. Alright, let me tackle the first one first.

Starting with the first inequality: 0 ≥ 54p - 144. Maybe I can rearrange this to solve for p. Let me add 144 to both sides. That would give me 144 ≥ 54p. Wait, is that right? Let me check. If I have 0 ≥ 54p - 144, then adding 144 to both sides gives 144 ≥ 54p. Yes, that seems correct. Then, to solve for p, I need to divide both sides by 54. So dividing both sides by 54, I get 144/54 ≥ p. Let me simplify that fraction. Both numerator and denominator are divisible by 18, right? 144 divided by 18 is 8, and 54 divided by 18 is 3. So that simplifies to 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 rearrange this one as well. Let me add 20p to both sides. That would give me 20p > 12. Wait, let me check. Starting with 0 > 12 - 20p. If I add 20p to both sides, I get 20p > 12. Yes, that's correct. Then, to solve for p, divide both sides by 20. So p > 12/20. Simplifying that fraction: 12 divided by 20. Both numerator and denominator can be divided by 4. 12 divided by 4 is 3, and 20 divided by 4 is 5. So that simplifies to p > 3/5. Okay, so the second inequality simplifies to p > 3/5.

So now, the problem asks for values of p that satisfy both inequalities. So p must be greater than 3/5 and less than or equal to 8/3. So in interval notation, that would be the interval from 3/5 to 8/3, including 8/3 but not including 3/5. Wait, let me confirm. Since the first inequality is p ≤ 8/3 and the second is p > 3/5, so combining them, it's 3/5 < p ≤ 8/3. So in interval notation, that would be (3/5, 8/3].

But let me check if I did everything correctly. Let me verify the inequalities again.

For the first inequality: 0 ≥ 54p - 144. Let me plug in p = 8/3. So 54*(8/3) - 144. 54 divided by 3 is 18, so 18*8 is 144. So 144 - 144 = 0. So 0 ≥ 0 is true. So p = 8/3 is included, which matches the first inequality.

For the second inequality: 0 > 12 - 20p. Let me plug in p = 3/5. So 12 - 20*(3/5) = 12 - (60/5) = 12 - 12 = 0. So 0 > 0 is false. Therefore, p = 3/5 is not included, which matches the second inequality. So the interval is indeed (3/5, 8/3].

But let me check another point to be sure. Let's take p = 1, which is between 3/5 (which is 0.6) and 8/3 (which is approximately 2.666...). So p = 1.

First inequality: 0 ≥ 54*1 - 144 = 54 - 144 = -90. 0 ≥ -90 is true. Second inequality: 0 > 12 - 20*1 = 12 - 20 = -8. 0 > -8 is true. So p = 1 is in the solution set.

What about p = 3/5? As we saw earlier, plugging into the second inequality gives 0 > 0, which is false. So p = 3/5 is excluded. What about p = 8/3? First inequality gives 0 ≥ 0, which is true. Second inequality: 0 > 12 - 20*(8/3). Let me calculate that. 20*(8/3) is 160/3. So 12 is 36/3. So 36/3 - 160/3 = -124/3. Thus 0 > -124/3 is true. Therefore, p = 8/3 is included.

Another test: p = 0. Let me see. If p = 0, first inequality: 0 ≥ 54*0 - 144 => 0 ≥ -144, which is true. But second inequality: 0 > 12 - 20*0 => 0 > 12, which is false. Therefore, p = 0 is not in the solution set, which is consistent with our interval (3/5, 8/3].

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

Another check: p = 2. Let's see. First inequality: 0 ≥ 54*2 - 144 = 108 - 144 = -36. True. Second inequality: 0 > 12 - 20*2 = 12 - 40 = -28. True. So p = 2 is in the solution set. Which is between 3/5 and 8/3.

So all these checks seem to confirm that the solution is (3/5, 8/3].

But just to make sure, let me write the steps again:

First inequality: 0 ≥ 54p - 144
Add 144: 144 ≥ 54p
Divide by 54: 144/54 ≥ p => simplifies to 8/3 ≥ p or p ≤ 8/3.

Second inequality: 0 > 12 - 20p
Subtract 12: -12 > -20p
Divide by -20 (remember to reverse inequality sign): (-12)/(-20) < p => 12/20 < p => simplifies to 3/5 < p.

Wait, hold on, let me check that step again. Because when solving inequalities, multiplying or dividing by a negative number reverses the inequality sign. So starting from 0 > 12 - 20p.

Let me solve it step by step. Starting with 0 > 12 - 20p. Let's subtract 12 from both sides: -12 > -20p. Then, divide both sides by -20. Since dividing by a negative number reverses the inequality sign, we get (-12)/(-20) < p. Which simplifies to 12/20 < p. Which simplifies to 3/5 < p. So that's correct. So p > 3/5.

Alternatively, another way to solve the second inequality: 0 > 12 - 20p. Let's add 20p to both sides: 20p > 12. Then divide by 20: p > 12/20 = 3/5. So same result.

Therefore, combining p > 3/5 and p ≤ 8/3. So the solution set is 3/5 < p ≤ 8/3.

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

But the problem says to express the answer in interval notation, reducing any fractions. So 3/5 is already reduced, and 8/3 is also reduced. So that should be the answer.

Wait, but maybe I should check if there's any mistake here. For example, could there be a case where the inequalities overlap differently? Let me think.

Alternatively, maybe I should graph both inequalities on a number line.

First inequality: p ≤ 8/3. So all numbers to the left of 8/3, including 8/3. Second inequality: p > 3/5. So all numbers to the right of 3/5, not including 3/5. The overlap is between 3/5 and 8/3, not including 3/5 and including 8/3. So yes, that's (3/5, 8/3].

Therefore, I think that is correct.

So the final answer is the interval from 3/5 to 8/3, with 3/5 excluded and 8/3 included.

I don't think I made any mistakes here. The key was to solve both inequalities correctly, remembering to reverse the inequality sign when dividing by a negative number. But in the second inequality, I didn't have to reverse the sign because when I added 20p, it was positive. Wait, actually when I solved it by adding 20p, I didn't have to reverse. But when I subtracted 12, I had -12 > -20p, which required reversing the inequality when dividing by -20. Either way, I got the same answer.

Therefore, confident with this solution.

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

To find all values of $ p $ that satisfy *both* the inequalities:

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

we begin by solving each inequality separately and then finding the intersection of the solution sets.

---

### **Step 1: Solve the first inequality**

$$
0 \geq 54p - 144
$$

Add 144 to both sides:

$$
144 \geq 54p
$$

Divide both sides by 54:

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

Simplify the fraction:

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

So:

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

---

### **Step 2: Solve the second inequality**

$$
0 > 12 - 20p
$$

Subtract 12 from both sides:

$$
-12 > -20p
$$

Divide both sides by $-20$, remembering that dividing by a **negative** number **flips** the inequality:

$$
\frac{-12}{-20} < p \Rightarrow \frac{12}{20} < p \Rightarrow \frac{3}{5} < p
$$

---

### **Step 3: Combine both inequalities**

Now we have:

- From the first inequality: $ p \leq \frac{8}{3} $
- From the second inequality: $ p > \frac{3}{5} $

So the values of $ p $ that satisfy **both** inequalities are those that are **strictly greater than** $ \frac{3}{5} $ **and at most** $ \frac{8}{3} $. That is:

$$
\frac{3}{5} < p \leq \frac{8}{3}
$$

---

### **Final Answer in Interval Notation**

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

[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. Let me write them down again to make sure I have them right.

First inequality: 0 ≥ 54p - 144. Second inequality: 0 > 12 - 20p. Alright, let me tackle the first one first.

Starting with the first inequality: 0 ≥ 54p - 144. Maybe I can rearrange this to solve for p. Let me add 144 to both sides. That would give me 144 ≥ 54p. Wait, is that right? Let me check. If I have 0 ≥ 54p - 144, then adding 144 to both sides gives 144 ≥ 54p. Yes, that seems correct. Then, to solve for p, I need to divide both sides by 54. So dividing both sides by 54, I get 144/54 ≥ p. Let me simplify that fraction. Both numerator and denominator are divisible by 18, right? 144 divided by 18 is 8, and 54 divided by 18 is 3. So that simplifies to 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 rearrange this one as well. Let me add 20p to both sides. That would give me 20p > 12. Wait, let me check. Starting with 0 > 12 - 20p. If I add 20p to both sides, I get 20p > 12. Yes, that's correct. Then, to solve for p, divide both sides by 20. So p > 12/20. Simplifying that fraction: 12 divided by 20. Both numerator and denominator can be divided by 4. 12 divided by 4 is 3, and 20 divided by 4 is 5. So that simplifies to p > 3/5. Okay, so the second inequality simplifies to p > 3/5.

So now, the problem asks for values of p that satisfy both inequalities. So p must be greater than 3/5 and less than or equal to 8/3. So in interval notation, that would be the interval from 3/5 to 8/3, including 8/3 but not including 3/5. Wait, let me confirm. Since the first inequality is p ≤ 8/3 and the second is p > 3/5, so combining them, it's 3/5 < p ≤ 8/3. So in interval notation, that would be (3/5, 8/3].

But let me check if I did everything correctly. Let me verify the inequalities again.

For the first inequality: 0 ≥ 54p - 144. Let me plug in p = 8/3. So 54*(8/3) - 144. 54 divided by 3 is 18, so 18*8 is 144. So 144 - 144 = 0. So 0 ≥ 0 is true. So p = 8/3 is included, which matches the first inequality.

For the second inequality: 0 > 12 - 20p. Let me plug in p = 3/5. So 12 - 20*(3/5) = 12 - (60/5) = 12 - 12 = 0. So 0 > 0 is false. Therefore, p = 3/5 is not included, which matches the second inequality. So the interval is indeed (3/5, 8/3].

But let me check another point to be sure. Let's take p = 1, which is between 3/5 (which is 0.6) and 8/3 (which is approximately 2.666...). So p = 1.

First inequality: 0 ≥ 54*1 - 144 = 54 - 144 = -90. 0 ≥ -90 is true. Second inequality: 0 > 12 - 20*1 = 12 - 20 = -8. 0 > -8 is true. So p = 1 is in the solution set.

What about p = 3/5? As we saw earlier, plugging into the second inequality gives 0 > 0, which is false. So p = 3/5 is excluded. What about p = 8/3? First inequality gives 0 ≥ 0, which is true. Second inequality: 0 > 12 - 20*(8/3). Let me calculate that. 20*(8/3) is 160/3. So 12 is 36/3. So 36/3 - 160/3 = -124/3. Thus 0 > -124/3 is true. Therefore, p = 8/3 is included.

Another test: p = 0. Let me see. If p = 0, first inequality: 0 ≥ 54*0 - 144 => 0 ≥ -144, which is true. But second inequality: 0 > 12 - 20*0 => 0 > 12, which is false. Therefore, p = 0 is not in the solution set, which is consistent with our interval (3/5, 8/3].

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

Another check: p = 2. Let's see. First inequality: 0 ≥ 54*2 - 144 = 108 - 144 = -36. True. Second inequality: 0 > 12 - 20*2 = 12 - 40 = -28. True. So p = 2 is in the solution set. Which is between 3/5 and 8/3.

So all these checks seem to confirm that the solution is (3/5, 8/3].

But just to make sure, let me write the steps again:

First inequality: 0 ≥ 54p - 144
Add 144: 144 ≥ 54p
Divide by 54: 144/54 ≥ p => simplifies to 8/3 ≥ p or p ≤ 8/3.

Second inequality: 0 > 12 - 20p
Subtract 12: -12 > -20p
Divide by -20 (remember to reverse inequality sign): (-12)/(-20) < p => 12/20 < p => simplifies to 3/5 < p.

Wait, hold on, let me check that step again. Because when solving inequalities, multiplying or dividing by a negative number reverses the inequality sign. So starting from 0 > 12 - 20p.

Let me solve it step by step. Starting with 0 > 12 - 20p. Let's subtract 12 from both sides: -12 > -20p. Then, divide both sides by -20. Since dividing by a negative number reverses the inequality sign, we get (-12)/(-20) < p. Which simplifies to 12/20 < p. Which simplifies to 3/5 < p. So that's correct. So p > 3/5.

Alternatively, another way to solve the second inequality: 0 > 12 - 20p. Let's add 20p to both sides: 20p > 12. Then divide by 20: p > 12/20 = 3/5. So same result.

Therefore, combining p > 3/5 and p ≤ 8/3. So the solution set is 3/5 < p ≤ 8/3.

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

But the problem says to express the answer in interval notation, reducing any fractions. So 3/5 is already reduced, and 8/3 is also reduced. So that should be the answer.

Wait, but maybe I should check if there's any mistake here. For example, could there be a case where the inequalities overlap differently? Let me think.

Alternatively, maybe I should graph both inequalities on a number line.

First inequality: p ≤ 8/3. So all numbers to the left of 8/3, including 8/3. Second inequality: p > 3/5. So all numbers to the right of 3/5, not including 3/5. The overlap is between 3/5 and 8/3, not including 3/5 and including 8/3. So yes, that's (3/5, 8/3].

Therefore, I think that is correct.

So the final answer is the interval from 3/5 to 8/3, with 3/5 excluded and 8/3 included.

I don't think I made any mistakes here. The key was to solve both inequalities correctly, remembering to reverse the inequality sign when dividing by a negative number. But in the second inequality, I didn't have to reverse the sign because when I added 20p, it was positive. Wait, actually when I solved it by adding 20p, I didn't have to reverse. But when I subtracted 12, I had -12 > -20p, which required reversing the inequality when dividing by -20. Either way, I got the same answer.

Therefore, confident with this solution.

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

To find all values of $ p $ that satisfy *both* the inequalities:

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

we begin by solving each inequality separately and then finding the intersection of the solution sets.

---

### **Step 1: Solve the first inequality**

$$
0 \geq 54p - 144
$$

Add 144 to both sides:

$$
144 \geq 54p
$$

Divide both sides by 54:

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

Simplify the fraction:

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

So:

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

---

### **Step 2: Solve the second inequality**

$$
0 > 12 - 20p
$$

Subtract 12 from both sides:

$$
-12 > -20p
$$

Divide both sides by $-20$, remembering that dividing by a **negative** number **flips** the inequality:

$$
\frac{-12}{-20} < p \Rightarrow \frac{12}{20} < p \Rightarrow \frac{3}{5} < p
$$

---

### **Step 3: Combine both inequalities**

Now we have:

- From the first inequality: $ p \leq \frac{8}{3} $
- From the second inequality: $ p > \frac{3}{5} $

So the values of $ p $ that satisfy **both** inequalities are those that are **strictly greater than** $ \frac{3}{5} $ **and at most** $ \frac{8}{3} $. That is:

$$
\frac{3}{5} < p \leq \frac{8}{3}
$$

---

### **Final Answer in Interval Notation**

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

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