Question

Let a line with direction ratios $a,-4a,-7$ be perpendicular to the lines with direction ratios $3,-1,2b$ 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

Answer: 10

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