If $x \cdot a = 0, x \cdot b = 0$ and $x \cdot c = 0$ for some non-zero vector $x$,then the true statement is

$[(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})] \cdot \vec{d} = \dots$

If $\overline{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}$,$\overline{b}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k}$ and $\overline{c}=3 \hat{\imath}+\lambda \hat{\jmath}+5 \hat{k}$ are coplanar,then $\lambda$ is the root of the equation

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:

Let $\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}$,$\vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$,and $\vec{u}$ be a vector such that $|\vec{u}|=\alpha > 0$. If the minimum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is $-\alpha \sqrt{3401}$,and $|\vec{u} \cdot \hat{i}|^2=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers,then $m + n$ is equal to $.........$.

If $\bar{a}=2 \hat{i}-\hat{j}+\hat{k}, \bar{b}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\bar{c}=3 \hat{i}+\lambda \hat{j}+5 \hat{k}$ are coplanar,then $\lambda$ is the root of the equation