If $|\vec{c}|^2 = 60$ and $\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$,then a value of $\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$ is

If $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$,then $\vec{a} + \vec{b} = \dots$

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:

The vectors $\overrightarrow{AB} = 3 \hat{i} + 4 \hat{k}$ and $\overrightarrow{AC} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $ABC$. The length of the median through $A$ is

$A$ vector $\vec{a}$ has components $2p$ and $1$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If,with respect to the new system,$\vec{a}$ has components $p+1$ and $1$,then:

If $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-3 \hat{j}-5 \hat{k}$ are the position vectors of the points $A$ and $B$ respectively, $C$ divides $AB$ in the ratio $2:3$ and $M$ is the mid-point of $AB$, then $5(\text{position vector of } C) - 2(\text{position vector of } M) =$