Question

If equation of the plane through the straight line $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5}$ and perpendicular to the plane $x-y+z+2=0$is$ax-by+cz+4=0,$ then find the value of $a^2+b^2+c$

• A. $12$
• B. $14$
• C. $16$
• D. $18$

Answer 1

Answer: (B)

$14$

Let equation of a plane containing the line be $\ell (x-1)+m(y+2)+nz=0$
then $2\ell -3m+5n=0$ and $\ell -m+n=0$
$\therefore \frac{\ell }{2}=\frac{m}{3}=\frac{n}{1}$
$\therefore$ the plane is$2(x-1)+3(y+2)+z=0$
i.e. $2x+3y+z+4=0$

$\therefore a=2,b=-3,c=1$
$\therefore a^2+b^2+c=14$