If the vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$,and $\vec{c}=3 \hat{i}+p \hat{j}+5 \hat{k}$ are coplanar,then $p=$

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:

The maximum volume of a parallelepiped (in cubic units) with vectors $\vec{u} = 2a\hat{i} + \hat{k}$, $\vec{v} = a\hat{j} - a\hat{k}$, and $\vec{w} = 3\hat{i} + a\hat{j}$, where $a \in [0, 1]$, as its coterminous edges is:

The scalar $\overline{a} \cdot [(\overline{b} + \overline{c}) \times (\overline{a} + \overline{b} + \overline{c})]$ equals

If $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\vec{a}|=2$,$|\bar{b}|=3$,$|\bar{c}|=4$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{3}$,then $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=$ (in $\sqrt{3}$)

If $\bar{a} = \bar{i} - \bar{j}$,$\bar{b} = \bar{j} - \bar{k}$,$\bar{c} = \bar{k} - \bar{i}$ and $\bar{d}$ is a unit vector such that $\bar{a} \cdot \bar{d} = 0$ and $[\bar{b} \bar{c} \bar{d}] = 0$,then the vector $\bar{d} = ....$