MHTCET 17 Apr 2026 Shift 2 Paper

Concept:The angle between a line and a plane is obtained using the sine of the angle between the line's direction vector and the plane's normal vector.Explanation:Let the common value of the given line equation be $\lambda$.So, $x-1=2-y=\frac{2z-6}{4}=\lambda$.This gives $x=1+\lambda$, $y=2-\lambda$, and $z=3+2\lambda$.Therefore, the direction vector of the line is $\vec{d}=\hat{i}-\hat{j}+2\hat{k}$.The normal vector of the plane is $\vec{n}=2\hat{i}+\hat{j}+\hat{k}$.If $\theta$ is the angle between the line and the plane, then $\sin\theta=\frac{|\vec{d}\cdot\vec{n}|}{|\vec{d}||\vec{n}|}$.Now, $\vec{d}\cdot\vec{n}=2-1+2=3$.Also, $|\vec{d}|=\sqrt{1+1+4}=\sqrt{6}$ and $|\vec{n}|=\sqrt{4+1+1}=\sqrt{6}$.Thus, $\sin\theta=\frac{3}{\sqrt{6}\cdot\sqrt{6}}=\frac{3}{6}=\frac{1}{2}$.Hence, $\theta=\frac{\pi}{6}$.Answer:The correct option is B: $\frac{\pi}{6}$.