If $a, b, c, d$ are $4$ vectors such that $a \cdot b = 0$,$|a \times c| = |a||c|$,and $|a \times d| = |a||d|$,then $[b c d] = $

Let $\vec{c}$ be a vector coplanar with the unit vectors $\vec{a}$ and $\vec{b}$,and let $\vec{d}$ be the unit vector perpendicular to $\vec{a}$,$\vec{b}$,and $\vec{c}$. If $[\vec{a} \vec{b} \vec{d}] \vec{c} - [\vec{a} \vec{b} \vec{c}] \vec{d} = \hat{i} + 2\hat{j} + 2\hat{k}$ and the angle between $\vec{a}$ and $\vec{b}$ is $30^{\circ}$,then $|\vec{c}| =$

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:

If $[\bar{a} \bar{b} \bar{c}] = 4$,then the volume (in cubic units) of the parallelepiped with $\bar{a} + 2 \bar{b}, \bar{b} + 2 \bar{c}$ and $\bar{c} + 2 \bar{a}$ as coterminal edges,is

If $2 \hat{i}-\hat{j}+3 \hat{k}$,$-12 \hat{i}-\hat{j}-3 \hat{k}$,$-\hat{i}+2 \hat{j}-4 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}-\hat{k}$ are the position vectors of four coplanar points,then $\lambda=$

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\vec{d}=3 \hat{i}+2 \hat{j}+6 \hat{k}$,and $|\vec{a}|=\sqrt{10}$. Then a possible value of $[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{d}]+[\vec{a} \vec{c} \vec{d}]$ is equal to: