If $P(1, 3, -7)$ and $Q(5, -2, 4)$,then $|\vec{PQ}| = \dots$

If $\vec{r} = 3\hat{i} + 2\hat{j} - 5\hat{k}$,$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$,and $\vec{c} = -2\hat{i} + \hat{j} - 3\hat{k}$ such that $\vec{r} = \lambda\vec{a} + \mu\vec{b} + \gamma\vec{c}$,then -

If the points whose position vectors are $2 \hat{i}+\hat{j}+\hat{k}$, $6 \hat{i}-\hat{j}+2 \hat{k}$, and $14 \hat{i}-5 \hat{j}+p \hat{k}$ are collinear, then the value of $p$ is:

Let $ABC$ be a triangle whose circumcentre is at $P$. If the position vectors of $A, B, C$ and $P$ are $\vec{a}, \vec{b}, \vec{c}$ and $\frac{\vec{a} + \vec{b} + \vec{c}}{4}$ respectively,then the position vector of the orthocentre of this triangle is:

If $\bar{x}$ is a non-zero vector and $k > 0, k \neq 1$,then $\frac{-k \bar{x}}{|\bar{x}|}$ is $.......$ .

If $2 \hat{i}+4 \hat{j}-5 \hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{j}+2 \hat{k}$ are the position vectors of the vertices $A$, $B$, and $C$ of a triangle respectively, then a unit vector along the median drawn through the vertex $A$ is