JEE Main 2014 — Three Dimensional Geometry Question with Solution

Given two planes: $$x-ay-b=0\text{ and }cy-z+d=0$$ Let, $l,m,n$ be the direction ratio of the required line. Since the required line is perpendicular to normal of both the plane, therefore $l-am=0$ and $cm-n=0$ $\Rightarrow l-am+0.n=0$ and $0.l+cm-n=0$ $$\therefore \frac{l}{a-0}=\frac{m}{0+1}=\frac{n}{c-0}$$ Hence, d.R of the required line are $a,1$, $c$. Hence, options (c) and (d) are rejected. Now, the point $(a+b,1,c+d)$ satisfy the equation of the two given planes. $\therefore$ Option (b) is correct.