MHTCET 18 Apr 2026 Shift 1 Paper

Concept:For a line to be parallel to a plane, its direction vector must be perpendicular to the normal vector of the plane.Explanation:Let the common ratio of the given line be $t$.A general point on the given line is $Q(-2+3t,\,-1-t,\,2+t)$.Line $\ell$ passes through $P(2,1,1)$ and intersects the given line at $Q$.So the direction vector of $\ell$ is $\overrightarrow{PQ}=(-4+3t,\,-2-t,\,1+t)$.The plane $x+y+2z=18$ has normal vector $\vec n=(1,1,2)$.Since $\ell$ is parallel to the plane, $\overrightarrow{PQ}\cdot \vec n=0$.Thus $(-4+3t)(1)+(-2-t)(1)+(1+t)(2)=0$.Simplifying gives $4t-4=0$, so $t=1$.Substituting $t=1$, we get $Q=(1,-2,3)$.Therefore $\overrightarrow{PQ}=(1-2,\,-2-1,\,3-1)=(-1,-3,2)$.Equivalently, direction ratios can be taken as $(1,3,-2)$.So the equation of $\ell$ is $\frac{x-2}{1}=\frac{y-1}{3}=\frac{z-1}{-2}$.This line also passes through $(3,4,-1)$, since $\frac{2-3}{1}=\frac{1-4}{3}=\frac{1+1}{-2}=-1$.Hence it matches option D.Answer:Option D: $\frac{x-3}{1}=\frac{y-4}{3}=\frac{z+1}{-2}$