If $l\vec{a} + m\vec{b} + n\vec{c} = \vec{0},$ where $l, m, n$ are scalars and $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular non-zero vectors,then

Let the position vectors of the vertices of a triangle $ABC$ be $\bar{a}, \bar{b}, \bar{c}$. If on the plane of the triangle,$P$ is a point having position vector $\bar{x}$ such that $\bar{x} \cdot (\bar{c} - \bar{b}) = \bar{a} \cdot \bar{c} - \bar{a} \cdot \bar{b}$ and $\bar{x} \cdot (\bar{a} - \bar{c}) = \bar{a} \cdot \bar{b} - \bar{b} \cdot \bar{c}$,then for the triangle $ABC$,$P$ is the

If $c = 2 \lambda (a \times b) + 3 \mu (b \times a)$ where $a \times b \neq 0$ and $c \cdot (a \times b) = 0$,then:

Two adjacent sides of a parallelogram $ABCD$ are given by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $\vec{AD}$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $\vec{AD}$ becomes $\vec{AD'}$. If $\vec{AD'}$ makes a right angle with the side $\vec{AB}$, then the cosine of the angle $\alpha$ is given by

If $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors and vector $\vec{c}$ is such that $\vec{c} = \vec{a} + \vec{b}$,then $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b})$ is

If $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$ and $|\vec{a} \times \vec{b}| = |\vec{a} \cdot \vec{b}|$, then $\theta$ is equal to