Question

Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$ 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

Answer: 355

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