Find the point of intersection of the lines
`(x+1)/(-3)=(y-3)/(2)=(z+2)/(1)and(x)/(1)=(y-7)/(-3)=(z+7)/(2)`

Options

A. (-2,1,3)
B. (2,1, -3)
C. (2,3,1)
D. (2,-1,3)

Answer

To find the point of intersection of the given lines, we will first express each line in parametric form and then set the corresponding coordinates equal to each other.

### Step 1: Parametric Equations of the First Line
The first line is given by:
\[ \frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1} \]
Let’s set this equal to a parameter \( \lambda \):
- From \( \frac{x + 1}{-3} = \lambda \):
\[ x = -3\lambda - 1 \]
- From \( \frac{y - 3}{2} = \lambda \):
\[ y = 2\lambda + 3 \]
- From \( \frac{z + 2}{1} = \lambda \):
\[ z = \lambda - 2 \]

So, the parametric equations for the first line are:
\[ (x, y, z) = (-3\lambda - 1, 2\lambda + 3, \lambda - 2) \]

### Step 2: Parametric Equations of the Second Line
The second line is given by:
\[ \frac{x}{1} = \frac{y - 7}{-3} = \frac{z + 7}{2} \]
Let’s set this equal to a parameter \( \mu \):
- From \( \frac{x}{1} = \mu \):
\[ x = \mu \]
- From \( \frac{y - 7}{-3} = \mu \):
\[ y = -3\mu + 7 \]
- From \( \frac{z + 7}{2} = \mu \):
\[ z = 2\mu - 7 \]

So, the parametric equations for the second line are:
\[ (x, y, z) = (\mu, -3\mu + 7, 2\mu - 7) \]

### Step 3: Setting the Coordinates Equal
For the lines to intersect, their coordinates must be equal:
1. From the x-coordinates:
\[ -3\lambda - 1 = \mu \quad \text{(1)} \]
2. From the y-coordinates:
\[ 2\lambda + 3 = -3\mu + 7 \quad \text{(2)} \]
3. From the z-coordinates:
\[ \lambda - 2 = 2\mu - 7 \quad \text{(3)} \]

### Step 4: Solving the Equations
**From equation (1)**, we can express \( \mu \) in terms of \( \lambda \):
\[ \mu = -3\lambda - 1 \]

**Substituting \( \mu \) into equation (2)**:
\[ 2\lambda + 3 = -3(-3\lambda - 1) + 7 \]
\[ 2\lambda + 3 = 9\lambda + 3 + 7 \]
\[ 2\lambda + 3 = 9\lambda + 10 \]
Rearranging gives:
\[ 2\lambda - 9\lambda = 10 - 3 \]
\[ -7\lambda = 7 \implies \lambda = -1 \]

**Now substituting \( \lambda = -1 \) back to find \( \mu \)**:
\[ \mu = -3(-1) - 1 = 3 - 1 = 2 \]

### Step 5: Finding the Point of Intersection
Now we substitute \( \lambda = -1 \) into the parametric equations of the first line:
\[ x = -3(-1) - 1 = 3 - 1 = 2 \]
\[ y = 2(-1) + 3 = -2 + 3 = 1 \]
\[ z = -1 - 2 = -3 \]

Thus, the point of intersection is:
\[ (2, 1, -3) \]

### Final Answer
The point of intersection of the lines is \( (2, 1, -3) \).