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:

Let $a, b$ and $c$ be distinct positive numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}$,$\hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ are coplanar,then $c$ is equal to:

If $a=2u+3v+7w$, $b=u+v-2w$ and $c=-u-2v-3w$, then $\left|\frac{[u, v, w]}{[a, b, c]}\right|(a+b+c) = $

If $a$ and $b$ are unit vectors perpendicular to each other, then $[a + (a \times b) \quad b + (a \times b) \quad (a \times b)]$ =

If $3 \hat{i}+3 \hat{j}+\sqrt{3} \hat{k}$,$\hat{i}+\hat{k}$,and $\sqrt{3} \hat{i}+\sqrt{3} \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda$ is equal to

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to: