Perpendicular 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.

Options

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

Answer

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 \( x, y, z \) in terms of \( \lambda \):
\[ x = 2\lambda - 2, \quad y = -\lambda - 1, \quad z = 3\lambda \]

### Step 2: Write the Point on the Line
Let \( K(\lambda) \) be the point on the line:
\[ K(\lambda) = (2\lambda - 2, -\lambda - 1, 3\lambda) \]

### Step 3: Find the Normal Vector of the Plane
The equation of the plane is:
\[ x + y + z = 3 \]
The normal vector \( \mathbf{n} \) of the plane can be derived from the coefficients of \( x, y, z \):
\[ \mathbf{n} = (1, 1, 1) \]

### Step 4: Find the Direction Ratios of the Perpendicular
The direction ratios of the perpendicular from point \( K(\lambda) \) to the plane will be the same as the normal vector of the plane, which are \( (1, 1, 1) \).

### Step 5: Parametrize the Perpendicular Line
The equation of the line passing through point \( K(\lambda) \) in the direction of \( \mathbf{n} \) can be written as:
\[ (x, y, z) = (2\lambda - 2, -\lambda - 1, 3\lambda) + t(1, 1, 1) \]
This gives us:
\[ x = 2\lambda - 2 + t, \quad y = -\lambda - 1 + t, \quad z = 3\lambda + t \]

### Step 6: Substitute into the Plane Equation
To find the foot of the perpendicular, we substitute these parametric equations into the plane equation:
\[ (2\lambda - 2 + t) + (-\lambda - 1 + t) + (3\lambda + t) = 3 \]
Simplifying this:
\[ (2\lambda - 2 - \lambda - 1 + 3\lambda + 3t) = 3 \]
\[ (4\lambda - 3 + 3t) = 3 \]
\[ 4\lambda + 3t = 6 \]
\[ 3t = 6 - 4\lambda \]
\[ t = \frac{6 - 4\lambda}{3} \]

### Step 7: Find the Coordinates of the Foot of the Perpendicular
Substituting \( t \) back into the parametric equations:
\[ x = 2\lambda - 2 + \frac{6 - 4\lambda}{3} \]
\[ y = -\lambda - 1 + \frac{6 - 4\lambda}{3} \]
\[ z = 3\lambda + \frac{6 - 4\lambda}{3} \]

### Step 8: Simplify the Coordinates
Now we simplify each coordinate:
1. For \( x \):
\[ x = 2\lambda - 2 + \frac{6 - 4\lambda}{3} = \frac{6\lambda - 6 + 6 - 4\lambda}{3} = \frac{2\lambda}{3} \]
2. For \( y \):
\[ y = -\lambda - 1 + \frac{6 - 4\lambda}{3} = \frac{-3\lambda - 3 + 6 - 4\lambda}{3} = \frac{-7\lambda + 3}{3} \]
3. For \( z \):
\[ z = 3\lambda + \frac{6 - 4\lambda}{3} = \frac{9\lambda + 6 - 4\lambda}{3} = \frac{5\lambda + 6}{3} \]

### Final Result
Thus, the coordinates of the foot of the perpendicular from the point on the line to the plane are:
\[ \left( \frac{2\lambda}{3}, \frac{-7\lambda + 3}{3}, \frac{5\lambda + 6}{3} \right) \]