MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Find the normal vector of the plane using the cross product of the line’s direction vector and a vector joining a known point on the line to the given point, then use the distance formula.Explanation:The given line is$$\vec{r}=(\hat{i}-3 \hat{j}+\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})$$So one point on the line is $A(1,-3,1)$ and its direction vector is$$\vec{d}=2 \hat{i}-2 \hat{j}+\hat{k}$$The plane also passes through $Q(1,0,-1)$.Since the plane contains the line and point $Q$, it contains the vector $\overrightarrow{AQ}$.$$\overrightarrow{AQ}=(1-1,\;0-(-3),\;-1-1)=(0,3,-2)$$The normal vector $\vec{n}$ is perpendicular to both $\vec{d}$ and $\overrightarrow{AQ}$, so take their cross product.$$\vec{n}=\vec{d}\times\overrightarrow{AQ} =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & -2 & 1\\ 0 & 3 & -2 \end{vmatrix}$$$$\vec{n}=\hat{i}(4-3)-\hat{j}(-4-0)+\hat{k}(6-0) =\hat{i}+4\hat{j}+6\hat{k}$$Thus, the normal vector is $(1,4,6)$ and the plane passes through $A(1,-3,1)$.Equation of the plane is$$1(x-1)+4(y+3)+6(z-1)=0$$$$\Rightarrow x-1+4y+12+6z-6=0$$$$\Rightarrow x+4y+6z+5=0$$Distance of this plane from the origin is$$\frac{|5|}{\sqrt{1^2+4^2+6^2}}=\frac{5}{\sqrt{53}}$$Comparing with $\frac{p}{\sqrt{53}}$, we get $p=5$.Answer:$p=5$, which is option C.