The value of $a$,so that the volume of the parallelepiped formed by $\hat{i} + a \hat{j} + \hat{k}$,$\hat{j} + a \hat{k}$,and $a \hat{i} + \hat{k}$ becomes minimum is

If $\vec{u} = \hat{i} + 2\hat{j} - 2\hat{k}$, $\vec{v} = 2\hat{i} + \hat{k}$ and $\vec{w}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

If the volume of a tetrahedron whose vertices are $A \equiv (1, -6, 10)$,$B \equiv (-1, -3, 7)$,$C \equiv (5, -1, k)$,and $D \equiv (7, -4, 7)$ is $11$ cubic units,then the value of $k$ is:

Three vectors $\hat{i}-\hat{k}$,$\lambda \hat{i}+\hat{j}+(1-\lambda) \hat{k}$,and $\mu \hat{i}+\lambda \hat{j}+(1+\lambda-\mu) \hat{k}$ represent the coterminous edges of a parallelepiped. The volume of the parallelepiped depends on:

Let $a, b$ and $c$ be three vectors. Then the scalar triple product $[a, b, c]$ is equal to:

Let $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$, $\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$. If $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$, then $\lambda$ is equal to