Question

Let a line with direction ratios $a ,-4 a ,-7$ be perpendicular to the lines with direction ratios $3,-1,2 b$ and $b, a,-2$. If the point of intersection of the line $\frac{ x +1}{ a ^2+ b ^2}=\frac{ y -2}{ a ^2- b ^2}=\frac{ z }{1}$ and the plane $x-y+z=0$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

Answer 1

$ (a,-4 a,-7) \perp \text { to }(3,-1,2 b) $
$ a=2 b$...(i)
$ (a,-4 a,-7) \perp \text { to }(b, a,-2) $
$ 3 a+4 a-14 b=0$
$a b-4 a^2+14=0$....(ii)
From Equations (i) and (ii)
$ 2 b^2-16 b^2+14=0$
$ b^2=1$
$ a^2=4 b^2=4 $
$ \frac{x+1}{5}=\frac{y-2}{3}=\frac{z}{1}=k $
$ \alpha=5 k-1, \beta=3 k+2, \gamma=k$
As $(\alpha, \beta, \gamma)$ satisfies $x-y+z=0$
$ 5 k -1-(3 k +2)+ k =0 $
$ k =1$
$ \therefore \alpha+\beta+\gamma=9 k +1=10$