If $a \cdot i = 4$,then $(a \times j) \cdot (2j - 3k) = $

If $7 \hat{i}-4 \hat{j}+5 \hat{k}$ is the position vector of the vertex $A$ of a tetrahedron $ABCD$ and $-\hat{i}+4 \hat{j}-3 \hat{k}$ is the position vector of the centroid of the triangle $BCD$,then the position vector of the centroid of the tetrahedron $ABCD$ is

If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \beta\hat{k}$ are mutually orthogonal,then $(\alpha, \beta) = $

If $\vec{a}=\frac{3}{2} \hat{k}$ and $\vec{b}=\frac{2 \hat{i}+2 \hat{j}-\hat{k}}{2}$,then the angle between $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is: (in $^{\circ}$)

Find the angle between the vectors $\hat{i}-2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$.

If $\triangle ABC$ is right-angled at $A$,where $A \equiv (4, 2, x)$,$B \equiv (3, 1, 8)$,and $C \equiv (2, -1, 2)$,then the value of $x$ is