MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Find the plane containing two parallel lines, then compute the perpendicular distance of the given point from that plane.Explanation:Both lines have the same direction vector $\vec{b}=(1,2,-2)$, so they are parallel.A point on the first line is $A=(0,4,-1)$ and a point on the second line is $B=(2,1,0)$.The vector joining these points is $\overrightarrow{AB}=B-A=(2,-3,1)$.A normal vector to the plane is given by the cross product:$\vec{n}=\vec{b}\times\overrightarrow{AB}=(1,2,-2)\times(2,-3,1)=(-4,-5,-7)$.Taking $\vec{n}=(4,5,7)$ as the normal, the plane through $A(0,4,-1)$ is:$4(x-0)+5(y-4)+7(z+1)=0 \Rightarrow 4x+5y+7z-13=0$.Distance of $P(2,5,10)$ from this plane is:$d=\frac{|4(2)+5(5)+7(10)-13|}{\sqrt{4^2+5^2+7^2}}=\frac{|8+25+70-13|}{\sqrt{90}}=\frac{90}{\sqrt{90}}$.Therefore, $d^{2}=\left(\frac{90}{\sqrt{90}}\right)^{2}=90$.Answer:$d^{2}=90$, which corresponds to option C.