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

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