If $a, b, c$ are non-coplanar vectors and $d = \lambda a + \mu b + \nu c$,then $\lambda$ is equal to

If $\vec{p}$ and $\vec{q}$ are unequal unit vectors such that $(\vec{p} - \vec{q}) \cdot ((2\vec{q} + \vec{p}) \times (3\vec{p} - \vec{q})) = |\vec{p} + \vec{q}|$,then the angle between $\vec{p}$ and $\vec{q}$ is:

If $\bar{a}=\hat{i}+5 \hat{k}, \bar{b}=2 \hat{i}+3 \hat{k}, \bar{c}=4 \hat{i}-\hat{j}+2 \hat{k}$ and $\bar{d}=\hat{i}-\hat{j}$,then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=$

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to:

If the four points with position vectors $-2\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, $\hat{j}-\hat{k}$, and $\lambda\hat{j}+\hat{k}$ are coplanar, then $\lambda$ is equal to

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