Line ` (x-2)/(-2) = (y+1)/(-3) = (z-5)/1 ` intersects the plane `3x+4y+z = 3 ` in the point

Options

A. `(23/7,29/7,38/7)`
B. `(29/7,38/7,23/7)`
C. `(26/17,-29/17,89/17)`
D. `(1,2,8)`

Answer

To find the point of intersection of the line given by

\[ \frac{x-2}{-2} = \frac{y+1}{-3} = \frac{z-5}{1} \]

with the plane given by

\[ 3x + 4y + z = 3, \]

we can follow these steps:

### Step 1: Parametrize the Line
We can express the line in parametric form by introducing a parameter \( k \):

\[ \frac{x-2}{-2} = k \implies x = -2k + 2, \]
\[ \frac{y+1}{-3} = k \implies y = -3k - 1, \]
\[ \frac{z-5}{1} = k \implies z = k + 5. \]

### Step 2: Substitute into the Plane Equation
Now we substitute \( x \), \( y \), and \( z \) into the plane equation \( 3x + 4y + z = 3 \):

\[ 3(-2k + 2) + 4(-3k - 1) + (k + 5) = 3. \]

### Step 3: Simplify the Equation
Expanding this gives:

\[ -6k + 6 - 12k - 4 + k + 5 = 3. \]

Combining like terms:

\[ -17k + 7 = 3. \]

### Step 4: Solve for \( k \)
Now, we can solve for \( k \):

\[ -17k = 3 - 7, \]
\[ -17k = -4 \implies k = \frac{4}{17}. \]

### Step 5: Find the Coordinates
Now we substitute \( k \) back into the parametric equations to find the coordinates of the intersection point:

1. For \( x \):
\[ x = -2\left(\frac{4}{17}\right) + 2 = -\frac{8}{17} + \frac{34}{17} = \frac{26}{17}. \]

2. For \( y \):
\[ y = -3\left(\frac{4}{17}\right) - 1 = -\frac{12}{17} - \frac{17}{17} = -\frac{29}{17}. \]

3. For \( z \):
\[ z = \frac{4}{17} + 5 = \frac{4}{17} + \frac{85}{17} = \frac{89}{17}. \]

### Final Answer
Thus, the point of intersection is

\[ \left(\frac{26}{17}, -\frac{29}{17}, \frac{89}{17}\right). \]