Line of Feet of Perpendiculars from a Given Line to a Plane

To find the line along which the feet of the perpendiculars lie, we need to find the orthogonal projection of the given line on the plane.

Step 1: Parametrize the given line The given line is: $\frac{x+2}{2} = \frac{y+1}{-1} = \frac{z}{3} = \lambda$

A general point $P(\lambda)$ on this line can be written in terms of parameter $\lambda$ as: $P(\lambda) = (2\lambda - 2, -\lambda - 1, 3\lambda)$

Step 2: Find the foot of the perpendicular on the plane The given plane is: $x + y + z = 3 \implies x + y + z - 3 = 0$

The normal vector to the plane is $\vec{n} = (1, 1, 1)$.

Let $F(x, y, z)$ be the foot of the perpendicular dropped from point $P(2\lambda - 2, -\lambda - 1, 3\lambda)$ onto the plane. Using the formula for the foot of the perpendicular: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{a x_1 + b y_1 + c z_1 + d}{a^2 + b^2 + c^2}$

Substituting the coordinates of $P(\lambda)$ and the plane equation: $\frac{x - (2\lambda - 2)}{1} = \frac{y - (-\lambda - 1)}{1} = \frac{z - 3\lambda}{1} = -\frac{(2\lambda - 2) + (-\lambda - 1) + 3\lambda - 3}{1^2 + 1^2 + 1^2}$ $= -\frac{4\lambda - 6}{3} = \frac{6 - 4\lambda}{3}$

Step 3: Express the coordinates of the foot $F$ $x = (2\lambda - 2) + \frac{6 - 4\lambda}{3} = \frac{6\lambda - 6 + 6 - 4\lambda}{3} = \frac{2\lambda}{3}$ $y = (-\lambda - 1) + \frac{6 - 4\lambda}{3} = \frac{-3\lambda - 3 + 6 - 4\lambda}{3} = 1 - \frac{7\lambda}{3}$ $z = 3\lambda + \frac{6 - 4\lambda}{3} = \frac{9\lambda + 6 - 4\lambda}{3} = 2 + \frac{5\lambda}{3}$

Step 4: Form the equation of the line Rearranging these parametric equations by letting $t = \frac{\lambda}{3}$: $x = 2t \implies \frac{x}{2} = t$ $y - 1 = -7t \implies \frac{y - 1}{-7} = t$ $z - 2 = 5t \implies \frac{z - 2}{5} = t$

Equating the parameters gives the equation of the line: $\frac{x}{2} = \frac{y-1}{-7} = \frac{z-2}{5}$

Therefore, the correct option is (D).