MHTCET 11 Apr 2026 Shift 1 Paper

Concept:Use the angle bisector theorem and the section formula to find the point where the internal bisector of $\angle A$ meets $BC$.Explanation:The given position vectors give the coordinates:$A(0,7,10)$, $B(-1,6,6)$, $C(-4,9,6)$.Let the angle bisector of $\angle A$ meet $BC$ at $D$.By the angle bisector theorem,$\frac{BD}{DC} = \frac{AB}{AC}$.Compute $AB$:$AB = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{1+1+16} = 3\sqrt{2}$.Compute $AC$:$AC = \sqrt{(-4-0)^2 + (9-7)^2 + (6-10)^2} = \sqrt{16+4+16} = 6$.Therefore,$BD:DC = 3\sqrt{2}:6 = \sqrt{2}:2$.So $D$ divides $BC$ internally in the ratio $\sqrt{2}:2$.Using the section formula,$D = \frac{2B + \sqrt{2}C}{2+\sqrt{2}}$.Substitute the coordinates:$D = \frac{2(-1,6,6) + \sqrt{2}(-4,9,6)}{2+\sqrt{2}}$.$D = \frac{(-2-4\sqrt{2},\,12+9\sqrt{2},\,12+6\sqrt{2})}{2+\sqrt{2}}$.Simplifying each coordinate gives:$x = 2-3\sqrt{2}$,$y = 3+3\sqrt{2}$,$z = 6$.Answer:The position vector of the required point is$(2-3\sqrt{2})\hat{i} + (3+3\sqrt{2})\hat{j} + 6\hat{k}$.Correct option: B.