Find the torque of the couple formed by forces $(9, -1, 2)$ and $(3, -2, 1)$ acting at the points $5\hat{i} + \hat{k}$ and $-5\hat{i} - \hat{k}$ respectively.

Consider three vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$. Let $|\overrightarrow{a}|=2, |\overrightarrow{b}|=3$ and $\overrightarrow{a}=\overrightarrow{b} \times \overrightarrow{c}$. If $\alpha \in [0, \frac{\pi}{3}]$ is the angle between the vectors $\overrightarrow{b}$ and $\overrightarrow{c}$,then the minimum value of $27|\overrightarrow{c}-\overrightarrow{a}|^2$ is equal to :

In a triangle $ABC$, $D$ and $E$ divide the sides $BC$ and $CA$ in the ratio $2:1$ respectively. If $P$ is the point of intersection of $AD$ and $BE$, then the ratio in which $P$ divides $AD$ is

If $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$, $\vec{x}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}$, $\vec{y}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then $x^2+y^2=$

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $|\vec{a}+3 \vec{b}-4 \vec{c}|^2=$

The projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv (2, -3, 0)$,$B \equiv (1, -4, -2)$,$C \equiv (4, 6, 8)$,and $D \equiv (7, 0, 10)$ is