The projection of the vector $i+j+k$ along the vector $j$ is

The two vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}+5\hat{k}$ represent the two sides $\vec{AB}$ and $\vec{AC}$ respectively of a $\triangle ABC$. The length of the median through $A$ is

Find the unit vector in the direction of vector $\overrightarrow{PQ},$ where $P$ and $Q$ are the points $(1, 2, 3)$ and $(4, 5, 6),$ respectively.

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{3}$ and $(\vec{a}+\vec{b}+\vec{c})^2+(\vec{b}+\vec{c}-\vec{a})^2+(\vec{c}+\vec{a}-\vec{b})^2=36$,then $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2=$

If $\vec{a}=-2 \hat{i}+9 \hat{j}-6 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}+6 \hat{k}$ are vectors such that $|\vec{a}+\vec{b}|=25$,then the sum of the values of $t$ is

Let $\overrightarrow{OA} = \hat{i} - 3\hat{j} + \hat{k}$, $\overrightarrow{OB} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\overrightarrow{OC} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ be the position vectors of three points $A$, $B$, and $C$. Let $P$ be the point which divides $AB$ in the ratio $2:1$. If $l, m, n$ are the direction cosines of the vector $\overrightarrow{PC}$, then $l + 3m + 2n =$