The point of intersection of the lines `(x-5)/3=(y-7)/(-1)=(z+2)/1a n d=(x+3)/(-36)=(y-3)/2=(z-6)/4` is a. `(21 ,5/3,(10)/3)` b. `(2,10 ,4)` c. `(-3,3,6)` d. `(5,7,-2)`

Options

A. `(21,(5)/(3),(10)/(3))`
B. (2,10,4)
C. (-3,3,6)
D. (5,7,-2)

Explanation

To find the point of intersection of the two lines given by their symmetric equations, we will convert them into parametric form and then solve for the parameters where the coordinates are equal.

### Step 1: Write the equations in parametric form

For the first line:
\[ \frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1} = \lambda \]
This gives us:
- \( x = 3\lambda + 5 \)
- \( y = -\lambda + 7 \)
- \( z = \lambda - 2 \)

For the second line:
\[ \frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4} = \mu \]
This gives us:
- \( x = -36\mu - 3 \)
- \( y = 2\mu + 3 \)
- \( z = 4\mu + 6 \)

### Step 2: Set the coordinates equal to find the parameters

We now have:
1. \( 3\lambda + 5 = -36\mu - 3 \)
2. \( -\lambda + 7 = 2\mu + 3 \)
3. \( \lambda - 2 = 4\mu + 6 \)

### Step 3: Rearranging the equations

From equation 1:
\[ 3\lambda + 36\mu = -8 \quad \text{(Equation 1)} \]

From equation 2:
\[ \lambda + 2\mu = 4 \quad \text{(Equation 2)} \]

From equation 3:
\[ \lambda - 4\mu = 8 \quad \text{(Equation 3)} \]

### Step 4: Solve the equations

We will solve equations 2 and 3 first.

From equation 2:
\[ \lambda = 4 - 2\mu \]

Substituting \(\lambda\) into equation 3:
\[ (4 - 2\mu) - 4\mu = 8 \]
\[ 4 - 6\mu = 8 \]
\[ -6\mu = 4 \implies \mu = -\frac{2}{3} \]

Now substituting \(\mu\) back into equation 2 to find \(\lambda\):
\[ \lambda = 4 - 2\left(-\frac{2}{3}\right) = 4 + \frac{4}{3} = \frac{12}{3} + \frac{4}{3} = \frac{16}{3} \]

### Step 5: Find the point of intersection

Now substitute \(\lambda = \frac{16}{3}\) and \(\mu = -\frac{2}{3}\) back into the parametric equations to find the coordinates:

For \(x\):
\[ x = 3\left(\frac{16}{3}\right) + 5 = 16 + 5 = 21 \]

For \(y\):
\[ y = -\left(\frac{16}{3}\right) + 7 = -\frac{16}{3} + \frac{21}{3} = \frac{5}{3} \]

For \(z\):
\[ z = \left(\frac{16}{3}\right) - 2 = \frac{16}{3} - \frac{6}{3} = \frac{10}{3} \]

Thus, the point of intersection is:
\[ \left(21, \frac{5}{3}, \frac{10}{3}\right) \]

### Final Answer
The point of intersection of the two lines is:
\[ \boxed{\left(21, \frac{5}{3}, \frac{10}{3}\right)} \]

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