JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Three Dimensional Geometry

The family of planes through the line of intersection is $(2x-y+z-3)+\lambda (4x-3y+5z+9)=0$, which simplifies to $(2+4\lambda )x-(1+3\lambda )y+(1+5\lambda )z+(-3+9\lambda )=0$. For the plane to be parallel to the line with direction ratios $(-2,4,5)$, the normal vector must be perpendicular to it: $-2(2+4\lambda )+4(-(1+3\lambda ))+5(1+5\lambda )=0$. Expanding: $-4-8\lambda -4-12\lambda +5+25\lambda =0$, so $5\lambda -3=0$ and $\lambda =3/5$. Substituting back: $(2+12/5)x-(1+9/5)y+(1+3)z+(-3+27/5)=0$, giving $\frac{22}{5}x-\frac{14}{5}y+4z+\frac{12}{5}=0$, or $11x-7y+10z+6=0$. Thus $a=11$, $b=-7$, $c=10$, and $a+b+c=14$.