Question

Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $a x+b y+3 z=2(a+b)$ and the distance of the plane $P$ from the point $(1,27,7)$ be $c$. Then $a^2+b^2+c^2$ is equal to ___

Answer 1

The line $\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}$ have a point $(-10,8,0)$ with d. r. $(1,-2,1)$
$ \because \text { the plane } a x+b y+3 z=2(a+b)$
$ \Rightarrow b=2 a$
$\&$ dot product of d.r.'s is zero
$\therefore a -2 b +3=0$
$\therefore a =1 \& b =2$
Distance from $(1,27,7)$ is
$c=\frac{1+54+21-6}{\sqrt{14}}=\frac{70}{\sqrt{14}}=5 \sqrt{14}$
$ \therefore a ^2+ b ^2+ c ^2=1+4+350$
$ =355$