Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{c} = 3\hat{i} - 2\hat{k}$. If a vector $\vec{p}$ satisfies the conditions $\vec{p} \cdot \vec{c} = 0$ and $\vec{p} \times \vec{a} = \vec{b} \times \vec{a}$, then the value of $|\vec{p}|$ is:

Let $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j} - 3\hat{k}$ be three given vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$ and $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,then $|\overrightarrow{r}|$ is equal to:

The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is

If $x + y + z = 0$,$|x| = |y| = |z| = 2$ and $\theta$ is the angle between $y$ and $z$,then the value of $\csc^2 \theta + \cot^2 \theta$ is equal to

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a}+\vec{b}$ is a unit vector if $\theta =$

Let $PQR$ be a triangle such that $\overrightarrow{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{PR}=a\hat{i}+b\hat{j}-4\hat{k}$, where $a, b \in \mathbb{Z}$. Let $S$ be the point on $QR$, which is equidistant from the lines $PQ$ and $PR$. If $|\overrightarrow{PR}|=9$ and $\overrightarrow{PS}=\hat{i}-7\hat{j}+2\hat{k}$, then the value of $3a-4b$ is . . . . . . .