MHTCET 22 Apr 2025 Shift 1 Paper

Concept:For a line to lie in a plane, its direction vector must be perpendicular to the plane's normal vector, and every point on the line must satisfy the plane equation.Explanation:Write the given line in parametric form using the parameter $t$.From $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}=t$, we get $x=2+3t$, $y=1-5t$, and $z=-2+2t$.At $t=0$, a point on the line is $P(2,1,-2)$.The direction vector of the line is $\vec{d}=(3,-5,2)$.The normal vector of the plane $x+3y-\alpha z+\beta=0$ is $\vec{n}=(1,3,-\alpha)$.Since the line lies in the plane, $\vec{d}$ must be perpendicular to $\vec{n}$, so their dot product is zero.Therefore, $3(1)+(-5)(3)+2(-\alpha)=0$.This simplifies to $3-15-2\alpha=0$, so $-12-2\alpha=0$, giving $\alpha=-6$.Now substitute the point $P(2,1,-2)$ into the plane equation with $\alpha=-6$.The plane becomes $x+3y+6z+\beta=0$.Substituting the point gives $2+3(1)+6(-2)+\beta=0$.Thus, $2+3-12+\beta=0$, which simplifies to $-7+\beta=0$, so $\beta=7$.Finally, $\beta-\alpha=7-(-6)=13$.Answer:$(\beta-\alpha)=13$, so the correct option is B.