In a triangle $PQR$,let $\vec{a}=\vec{QR}, \vec{b}=\vec{RP}$ and $\vec{c}=\vec{PQ}$. If $|\vec{a}|=3, |\vec{b}|=4$ and $\frac{\vec{a} \cdot(\vec{c}-\vec{b})}{\vec{c} \cdot(\vec{a}-\vec{b})}=\frac{|\vec{a}|}{|\vec{a}|+|\vec{b}|}$,then the value of $|\vec{a} \times \vec{b}|^2$ is.

$\vec{c}$ is a vector along the bisector of the internal angle between the vectors $\vec{a}=4 \hat{i}+7 \hat{j}-4 \hat{k}$ and $\vec{b}=12 \hat{i}-3 \hat{j}+4 \hat{k}$. If the magnitude of $\vec{c}$ is $3 \sqrt{13}$,then $\vec{c}=$

Let $A = \hat{i} + 2 \hat{j}$. If $B$ is a vector in the $XY$ plane such that $(A + B) \cdot B = 15$ and $A \cdot B = 6$,then $|B|$ is

Let $\overline{a}, \overline{b}$,and $\overline{c}$ be three non-zero vectors such that no two of these are collinear. If the vector $\overline{a}+2\overline{b}$ is collinear with $\overline{c}$ and $\overline{b}+3\overline{c}$ is collinear with $\overline{a}$,then $\overline{a}+2\overline{b}+6\overline{c}$ equals

Let $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$,$\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\vec{a}\times\vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11}$,$|\vec{c}\times\vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a}\cdot\vec{d}$ is equal to

The vector projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)$ and $D \equiv(7,0,10)$,is