The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

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

If $a \times r = b + \lambda a$ and $a \cdot r = 3,$ where $a = 2i + j - k$ and $b = -i - 2j + k,$ then $r$ and $\lambda$ are equal to

The position vectors of the vertices of a triangle $ABC$ are $4i - 2j$,$i + 4j - 3k$,and $-i + 5j + k$ respectively. Then $\angle ABC = $

If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 8$,then $|\vec{x}| = $ . . . . . . .

$\vec{c}$ 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 $\vec{c}=$