If $a, b,$ and $c$ are unit coplanar vectors,then the scalar triple product $[a, b, c]$ is equal to:

Let $\overrightarrow{a}=\hat{i}-2 \hat{j}$,$\overrightarrow{b}=2 \hat{j}+3 \hat{k}$,$\overrightarrow{c}=p\hat{i}+q \hat{j}$ and $\overrightarrow{d}=p \hat{j}-q \hat{k}$ be four vectors. If $(\vec{a} \times \vec{b}) \cdot \vec{c}=3=(\vec{a} \times \vec{b}) \cdot \vec{d}$,then $3 p+q=$

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

If $3 \hat{i}-2 \hat{j}-\hat{k}$,$2 \hat{i}+3 \hat{j}-4 \hat{k}$,$-\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are respectively the position vectors of four coplanar points $P, Q, R$ and $S$,then $\lambda=$

If the vectors $m \hat{i} + m \hat{j} + n \hat{k}$,$\hat{i} + \hat{k}$,and $n \hat{i} + n \hat{j} + p \hat{k}$ lie in a plane,then...

If the vectors $(1 - x)\hat i + \hat j + \hat k$,$\hat i + (1 - y)\hat j + \hat k$,and $\hat i + \hat j + (1 - z)\hat k$ are coplanar,then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is $(x, y, z \neq 0)$.