Three vectors of magnitudes $a, 2a, 3a$ are along the directions of the diagonals of $3$ adjacent faces of a cube that meet at a point. The magnitude of the sum of these vectors is: (in $a$)

In $\triangle ABC$,the points $P, Q, R$ divide $BC, CA, AB$ in the ratio $3:4, 2:5, 9:5$,respectively,and the point $D$ divides $BC$ in the ratio $2:3$. If $\vec{AP} + \vec{BQ} + \vec{CR} = k \vec{AD}$,then $(14k + 1) : (14k - 1) = $

If $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{a}=2 \hat{i}+2 \hat{j}+p \hat{k}$,$|\vec{b}|=7$,$\vec{a} \cdot \vec{b}=4$ and $|\vec{a} \times \vec{b}|=5 \sqrt{17}$,then $p=$

For any two non-zero vectors $\vec{a}$ and $\vec{b}$,$(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$ is equal to:

If $ 2 \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| $,then the angle between $ \vec{a} $ and $ \vec{b} $ is: (in $^{\circ}$)

If $a = i + j + k$,$a \cdot b = 1$ and $a \times b = j - k$,then $b = $