The point at which the line joining the points `(2, -3, 1) and (3, -4, -5)` intersects the plane `2x+y+z=7` is

Options

A. `(1, 2, 7)`
B. ` (1, -2, 7)`
C. ` (-1, 2, 7)`
D. `(1, -2, -7)`

Answer

To find the point at which the line joining the points \( (2, -3, 1) \) and \( (3, -4, -5) \) intersects the plane given by the equation \( 2x + y + z = 7 \), we can follow these steps:

### Step 1: Determine the direction vector of the line
The direction vector \( \mathbf{d} \) of the line joining the points \( (2, -3, 1) \) and \( (3, -4, -5) \) can be calculated as:
\[ \mathbf{d} = (3 - 2, -4 - (-3), -5 - 1) = (1, -1, -6) \]

### Step 2: Parametric equations of the line
Using the point \( (2, -3, 1) \) as the starting point, we can write the parametric equations of the line:
\[ x = 2 + t \\ y = -3 - t \\ z = 1 - 6t \]
where \( t \) is a parameter.

### Step 3: Substitute into the plane equation
We substitute the parametric equations into the plane equation \( 2x + y + z = 7 \):
\[ 2(2 + t) + (-3 - t) + (1 - 6t) = 7 \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ 4 + 2t - 3 - t + 1 - 6t = 7 \\ 4 - 3 + 1 + (2t - t - 6t) = 7 \\ 2 - 5t = 7 \]

### Step 5: Solve for \( t \)
Rearranging gives:
\[ -5t = 7 - 2 \\ -5t = 5 \\ t = -1 \]

### Step 6: Find the coordinates of the intersection point
Now, substitute \( t = -1 \) back into the parametric equations to find the coordinates:
\[ x = 2 + (-1) = 1 \\ y = -3 - (-1) = -2 \\ z = 1 - 6(-1) = 1 + 6 = 7 \]

### Final Answer
Thus, the point of intersection is:
\[ (1, -2, 7) \]