Animated Solution for Mathematics - Vector Algebra: Let $x_0$ be the point of local maxima of $f(x)=\vec{a}\cdot (\vec{b}\times \vec{c})$, where $\vec{a}=x\hat{i}-2\hat{j}+3\hat{k},\vec{b}=-2\hat{i}+x\hat{j}-\hat{k}$ and $\vec{c}=7\hat{i}-2\hat{j}+x\hat{k}$. Then the value of $\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}$ at $x=x_0$ is

The problem involves three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, whose components depend on a variable $x$. The scalar triple product of these vectors represents the volume of the parallelepiped they define, which we express as a function $f(x)$ using a determinant.

The determinant is defined as:

To find the explicit form of $f(x)$, we expand the determinant along the first row:

Simplifying the expression step-by-step, we obtain:

This cubic polynomial represents the volume of the geometric shape as a function of $x$.

To find the local maximum of the volume function, we calculate the first derivative:

Setting the derivative to zero to find the critical points:

We apply the second derivative test to classify these points, where $f^{\prime \prime }(x)=6x$:

At $x=3$, $f^{\prime \prime }(3)=18>0$, which indicates a local minimum.

At $x=-3$, $f^{\prime \prime }(-3)=-18<0$, which confirms a local maximum. Thus, our critical value is $x_0=-3$.

We are tasked with evaluating the final expression based on the derived critical point. Given the simplified relationship $3x-13$:

Substituting $x=-3$ into the expression:

The final result of the calculation is -22. This process demonstrates the transition from geometric vector representation to analytical calculus and algebraic evaluation.