$\vec{c}$ is a unit vector in the direction of the sum of vectors $\vec{a}$ and $\vec{b}$. Where,$\vec{a} = 2 \hat{i} + 2 \hat{j} - 5 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$,then $|\vec{c}| = $ . . . . . . .

In a parallelogram $ABCD$,if $\vec{AC}$ and $\vec{BD}$ are the diagonals,then which of the following is equal to $\vec{AC} + \vec{BD}$?

If $\bar{a}=2 \hat{i}-\hat{j}+\hat{k}$,$\bar{b}=\hat{i}+\hat{j}-2 \hat{k}$ and $\bar{c}=4 \hat{i}-2 \hat{j}+\hat{k}$,then the unit vector in the direction of $3 \bar{a}+\bar{b}-2 \bar{c}$ is

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.

If non-zero vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are related by $\vec{a} = 8\vec{b}$ and $\vec{c} = -7\vec{b}$,find the angle between $\vec{a}$ and $\vec{c}$.

$ABCD$ is a tetrahedron. $\bar{i}-2\bar{j}+3\bar{k}$, $-2\bar{i}+\bar{j}+3\bar{k}$, and $3\bar{i}+2\bar{j}-\bar{k}$ are the position vectors of the points $A, B, C$ respectively. $-\bar{i}+2\bar{j}-3\bar{k}$ is the position vector of the centroid of the triangular face $BCD$. If $G$ is the centroid of the tetrahedron, then $GD=$