MHTCET 15 May 2024 Shift 1 Paper

Concept:A plane containing two given lines is determined by a point on either line and the cross product of the two direction vectors, giving its normal vector. Then use the perpendicular distance formula of a point from a plane.Explanation:The direction vectors of the two lines are $\vec{d_1} = 3\hat{i}+5\hat{j}+7\hat{k}$ and $\vec{d_2} = \hat{i}+4\hat{j}+7\hat{k}$.Take a point on the second line, say $(1,4,-4)$.The equation of the plane through this point with these direction vectors is given by the determinant:$$\begin{vmatrix} x-1 & y-4 & z+4 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}=0$$Expanding the determinant:$$(x-1)(7) - (y-4)(14) + (z+4)(7) = 0$$Simplify:$$7x - 7 - 14y + 56 + 7z + 28 = 0$$$$7x - 14y + 7z + 77 = 0$$Divide throughout by $7$:$$x - 2y + z + 11 = 0$$The perpendicular distance from the origin $(0,0,0)$ to this plane is:$$\left|\frac{0 - 2(0) + 0 + 11}{\sqrt{1^2 + (-2)^2 + 1^2}}\right| = \frac{11}{\sqrt{6}}$$Answer:The perpendicular distance is $\frac{11}{\sqrt{6}}$ units.Therefore, the correct option is A. $\frac{11}{\sqrt{6}}$ units.