The Geometry of Perpendiculars

Let $\frac{x + 2}{2} = \frac{y + 1}{-1} = \frac{z}{3} = a$ , where $a$ is any constant.

Any point on the above line has:

$x = 2a - 2$ , $y = -a - 1$ and $z = 3a$

Let the foot of the perpendicular from $(2a - 2,-a - 1, 3a)$ to $x + y + z = 3$ be $(x_{1}, y_{1}, z_{1})$ .

$\implies$ $\frac{x_{1} - (2a - 2)}{1} = \frac{y_{1} - (-a - 1)}{1} = \frac{z_{1} - 3a}{1} = -\frac{2a - 2 - a - 1 + 3a - 3}{1 + 1 + 1}$

$\implies$ $\frac{x_{1} - 2a + 2}{1} = \frac{y_{1} + a + 1}{1} = \frac{z_{1} - 3a}{1} = \frac{-4a + 6}{3}$

$\implies$ $x_{1} - 2a + 2 = y_{1} + a + 1 = z_{1} - 3a = 2 - \frac{4a}{3}$

$\implies$ $x_{1} = \frac{2a}{3}, y_{1} = 1 - \frac{7a}{3} = 2 + \frac{5a}{3}$

$\implies$ $a = \frac{x_{1} - 0}{2/3} = \frac{y_{1} - 1}{-7/3} = \frac{z_{1} - 2}{5/3}$

$\implies$ $\boxed{\frac{a}{3} = \frac{x_{1}}{2} = \frac{y_{1} - 1}{-7} = \frac{z_{1} - 2}{5}}$