Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.

Given $ABC$ is an equilateral triangle of side length $1$ unit and $P$ be any arbitrary point on the circumcircle of triangle $ABC,$ then $|\vec{PA}|^2+|\vec{PB}|^2+|\vec{PC}|^2$ is equal to

Let $PQR$ be a triangle such that $\overrightarrow{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{PR}=a\hat{i}+b\hat{j}-4\hat{k}$,where $a, b \in \mathbb{Z}$. Let $S$ be the point on $QR$,which is equidistant from the lines $PQ$ and $PR$. If $|\overrightarrow{PR}|=9$ and $\overrightarrow{PS}=\hat{i}-7\hat{j}+2\hat{k}$,then the value of $3a-4b$ is . . . . . . .

If $\hat{a}, \hat{b}, \hat{c}$ are unit vectors,then the least value of $|\hat{a}+\hat{b}|^2+|\hat{b}+\hat{c}|^2+|\hat{c}+\hat{a}|^2$ will be-

If the angle between the vectors $\vec{a} = 2\lambda^2 \hat{i} + 4\lambda \hat{j} + \hat{k}$ and $\vec{b} = 7\hat{i} - 2\hat{j} + \lambda \hat{k}$ is obtuse,then the values of $\lambda$ lie in:

If $\bar{a}$ and $\bar{b}$ are mutually perpendicular unit vectors,then $(3\bar{a}+2\bar{b}) \cdot (5\bar{a}-6\bar{b}) = $