Find the angle between the diagonals of parallelogram $PQRS$,if $\vec{PQ} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{PS} = \hat{i} - 2\hat{k}$.

If $a \perp b$ and $(a+b) \perp (a+mb)$,then $m$ is equal to

Assertion $(A)$: $a, b, c, d$ are position vectors of $4$ points such that $2a - 3b + 7c - 6d = 0 \Rightarrow a, b, c, d$ are coplanar.
Reason $(R)$: Vector equation of the plane passing through three points whose position vectors are $a, b, c$ is $r = (1 - x - y)a + xb + yc$.
Which of the following is true?

If vectors $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$,and $\bar{c}=-3 \hat{i}+\hat{j}+2 \hat{k}$ are such that $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then $\lambda=$

Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}$,$\vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c}+\hat{i}) \times (\vec{a}+\vec{b}+\hat{i}) = \vec{a} \times (\vec{c}+\hat{i})$ and $\vec{a} \cdot \vec{c} = -29$. Then $\vec{c} \cdot (-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}|=7$,$|\vec{b}|=1$ and $|\vec{a} \times \vec{b}|^2 = k^2 - (\vec{a} \cdot \vec{b})^2$,then the values of $k$ and $\theta$ are