JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Three Dimensional Geometry

The plane through the line of intersection is $(x+2y+az-2)+\lambda (x-y+z-3)=0$, with coefficients proportional to those of $5x-11y+bz-(6a-1)=0$. Matching the $x$ and $y$ coefficients, $\frac{1+\lambda }{5}=\frac{2-\lambda }{-11}$ gives $\lambda =-\frac{7}{2}$ and the common ratio $\frac{1+\lambda }{5}=-\frac{1}{2}$. Matching the constant term, $\frac{2+3\lambda }{6a-1}=-\frac{1}{2}$ gives $-\frac{17}{2}=-\frac{1}{2}(6a-1)$, so $a=3$; matching the $z$ coefficient, $\frac{a+\lambda }{b}=-\frac{1}{2}$ gives $b=1$. The plane is $5x-11y+z=17$, and its distance from $(3,-c,c)$ is $\frac{|15+11c+c-17|}{\sqrt{147}}=\frac{|12c-2|}{7\sqrt{3}}$. Setting this equal to $\frac{2}{\sqrt{3}}$ gives $|12c-2|=14$, whose only integer solution is $c=-1$. Therefore $\frac{a+b}{c}=\frac{4}{-1}=-4$.