analytic geometry , Orthogonal projection

Question

I take the point $P = (2, 1, 0)$ , the line $r$ : $X = (0, 0, 0) + t(2, 1, 0)$ and the plane $B$: $x + y + z -3 = 0$. For each point $Q$ let it be $Q'$ its orthogonal projection to $B$. Find the $Q$ points that output the area of $\triangle PQQ' = \frac{9}{2}\sqrt{6}$

Answer 1

The point P lies in the plane B

The point Q (2t / t / 0)

has distance

$$QQ' = \frac{3t-3}{\sqrt{3}} = (t-1)\sqrt{3}$$

from the plane B

Intersecting the line

$$(2t,t,0) + r (1,1,1)$$

with the plane B gives

3t + 3r = 3 , so r = 1-t

So Q' (1+t , 1 , 1-t)

PQ' is the vector (t-1 , 0 , 1-t) , which has length $(t-1)\sqrt{2}$

Since PQ' and QQ' are orthogonal, the area of the triangle is

$$\frac{(t-1)^2\sqrt{6}}{2} = 4.5 * \sqrt{6}$$

So $(t-1)^2=9$

The solutions are t = 4 and t = -2.

So, the points Q are (8,4,0) and (-4,-2,0)