MHTCET 22 Apr 2025 Shift 2 Paper

Concept:A plane given in parametric form has normal vector equal to the cross product of its two direction vectors.Distance from origin is $\frac{|d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$ for plane $ax+by+cz+d=0$.Explanation:The plane passes through the point $(1,-1,0)$.Its direction vectors are $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-2\hat{j}+3\hat{k}$.Normal vector to the plane is their cross product:$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 1 & 1\\ 1 & -2 & 3 \end{vmatrix} = 5\hat{i}-2\hat{j}-3\hat{k}$$Using point $(1,-1,0)$, the plane equation is:$$5(x-1)-2(y+1)-3z=0$$$$\Rightarrow 5x-2y-3z=7$$Distance from origin $(0,0,0)$:$$d = \frac{|5(0)-2(0)-3(0)-7|}{\sqrt{5^{2}+(-2)^{2}+(-3)^{2}}} = \frac{7}{\sqrt{38}}$$Answer:$\frac{7}{\sqrt{38}}$ units, i.e. option A.