Perpendiculars are drawn from points on the line `(x+2)/2=(y+1)/(-1)=z/3` to the plane `x + y + z=3` The feet of perpendiculars lie on the line (a) `x/5=(y-1)/8=(z-2)/(-13)` (b) `x/2=(y-1)/3=(z-2)/(-5)` (c) `x/4=(y-1)/3=(z-2)/(-7)` (d) `x/2=(y-1)/(-7)=(z-2)/5`

Options

A. `(x)/(5)= (y-1)/(8) = (z-2)/(-13)`
B. `(x)/(2)= (y+1)/(3)=(z-2)/(-5)`
C. `(x)/(4)= (y-1)/(3)= (z-2)/(-7)`
D. `(x)/(2)= (y-1)/(-7)= (z-2)/(5)`

Explanation

To solve the problem, we need to find the feet of the perpendiculars drawn from points on the given line to the specified plane. Here’s a step-by-step solution:

### Step 1: Parametrize the Line
The line is given by the equation:
\[ \frac{x+2}{2} = \frac{y+1}{-1} = \frac{z}{3} \]
Let this common ratio be \( \lambda \). Then we can express the coordinates of any point on the line as:
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