MHTCET 11 Apr 2026 Shift 2 Paper

Concept:The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ can be written as a determinant of the three vectors. A local maximum occurs where $f'(x)=0$ and $f''(x)<0$.Explanation:Given $\vec{a}=x\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=-2\hat{i}+x\hat{j}-\hat{k}$, and $\vec{c}=7\hat{i}-2\hat{j}+x\hat{k}$, we form the determinant:$$f(x)=\begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$$Expanding along the first row:$$f(x)=x(x^{2}-2)+2(-2x+7)+3(4-7x)$$On simplification:$$f(x)=x^{3}-2x-4x+14+12-21x=x^{3}-27x+26$$For local maxima, set the first derivative to zero:$$f'(x)=3x^{2}-27=0 \Rightarrow x=\pm 3$$Now check the second derivative $f''(x)=6x$.At $x=-3$, $f''(-3)=-18<0$, so $x_{0}=-3$ is the point of local maxima.Now compute $\vec{a}\cdot\vec{c}$:$$\vec{a}\cdot\vec{c}=(7x)+4+(3x)=10x+4$$Substitute $x=x_{0}=-3$:$$\vec{a}\cdot\vec{c}=10(-3)+4=-30+4=-26$$Answer:The value of $\vec{a}\cdot\vec{c}$ at $x=x_{0}$ is $-26$.Hence, the correct option is D.