MHTCET 23 Apr 2025 Shift 1 Paper

Concept:For a plane containing a given line, the direction vector of the line lies in the plane. For two perpendicular planes, their normal vectors are perpendicular.Explanation:First, find the plane that contains the two lines with direction vectors $(2,3,1)$ and $(3,2,1)$.Its normal vector is the cross product:$$(2,3,1) \times (3,2,1) = (1,1,-5)$$So this plane is $x + y - 5z = 0$.Let the required plane be $ax + by + cz = 0$.Since it contains the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$, the direction vector $(1,2,3)$ lies in this plane:$$a + 2b + 3c = 0$$Since the required plane is perpendicular to $x + y - 5z = 0$, their normals are perpendicular:$$a + b - 5c = 0$$Subtract the second equation from the first:$$(a + 2b + 3c) - (a + b - 5c) = 0$$$$\Rightarrow b + 8c = 0 \Rightarrow b = -8c$$Substitute $b = -8c$ into $a + b - 5c = 0$:$$a - 8c - 5c = 0 \Rightarrow a = 13c$$Taking $c = 1$, we get $a = 13$ and $b = -8$.Hence the required plane is $13x - 8y + z = 0$.This matches option (C).Answer:$13x - 8y + z = 0$, i.e., option (C).