If \(x_0\) is the point of local minima of \(\mathrm{f}(x)=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})\) where \(\overline{\mathrm{a}}=x \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\), \(\overline{\mathrm{b}}=-2 \hat{\mathrm{i}}+x \hat{\mathrm{j}}-\hat{\mathrm{k}}\), \(\overline{\mathrm{c}}=7 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+x \hat{\mathrm{k}}\), then value of \(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}\) at \(x=x_0\) is

\(\begin{aligned} & \mathrm{f}(x)=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\ & =\left|\begin{array}{ccc}x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x\end{array}\right| \\ & =x\left(x^2-2\right)+2(-2 x+7)+3(4-7 x) \\ \therefore \quad & \mathrm{f}(x)=x^3-27 x+26 \\ & \text { Now, } \mathrm{f}^{\prime}(x)=0 \\ \Rightarrow & 3 x^2-27=0 \\ \Rightarrow & x^2=9 \\ \Rightarrow & x= \pm 3 \\ & \mathrm{f}^{\prime \prime}(x)=6 x \\ & \Rightarrow \mathrm{f}^{\prime \prime}(x)=18\gt0 \\ \therefore \quad & \mathrm{f}(x) \text { has local minimum at } x=3 .\end{aligned}\)
\(\begin{aligned} \therefore \quad \bar{a} & =3 \hat{i}-2 \hat{j}+3 \hat{k} \\ \bar{b} & =-2 \hat{i}+3 \hat{j}-\hat{k} \\ \therefore \quad \bar{a} \cdot \bar{b} & =(3 \hat{i}-2 \hat{j}+3 \hat{k}) \cdot(-2 \hat{i}+3 \hat{j}-\hat{k}) \\ & =3(-2)+(-2)(3)+3(-1) \\ & =-15\end{aligned}\)