QJEEAdv13Maaad0: JEE Advanced 2013 Mathematics - Class 12 JEEAdv13Maaad0 Solution: Perpendiculars are drawn from points on the line to the plane . …

Let $P$ be a general point on the line :

$\frac{x+2}{2} = \frac{y+1}{-1} = \frac{z}{3} = \lambda$

Therefore, the coordinates of $P$ can be expressed as :

$P = (-2 + 2\lambda, -1 - \lambda, 3\lambda)$

Now, let $(h, k, w)$ represent the foot of the perpendicular from $P$ to the plane $x + y + z = 3$ .

We use the formula for the foot of the perpendicular from a point to a plane :

$h = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + (-2 + 2\lambda), \\ k = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + (-1 - \lambda), \\ w = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + 3\lambda$

Simplifying these, we get :

$h = \frac{+2\lambda}{3}, \\ k = \frac{-7\lambda}{3} + 1, \\ w = \frac{5\lambda}{3} + 2$

So, we can express $\lambda$ and the coordinates $(h, k, w)$ as :

$\frac{h}{2} = \frac{k - 1}{-7} = \frac{w - 2}{5} = \frac{\lambda}{3}$

Therefore, the locus of the foot of the perpendiculars is :

$\frac{x}{2} = \frac{y - 1}{-7} = \frac{z - 2}{5} = \frac{\lambda}{3}$

Thus, the line on which the feet of the perpendiculars lie is represented as :

$\frac{x}{2} = \frac{y - 1}{-7} = \frac{z - 2}{5}$