Three vectors of magnitudes $a, 2a, 3a$ are along the directions of the diagonals of $3$ adjacent faces of a cube that meet at a point. The magnitude of the sum of these vectors is: (in $a$)

$A$ force of magnitude $5$ units acting along the vector $2i - 2j + k$ displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$. The work done is:

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

$A$ parallelogram is constructed with $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$ as its adjacent sides, where $|\vec{a}| = 2\sqrt{2}$ and $|\vec{b}| = 3$. The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. Find the lengths of the diagonals of the parallelogram.

The area of the rectangle having vertices $P, Q, R, S$ with position vectors $-\hat{i}+\hat{j}+\hat{k}, \hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{j}+\hat{k}, -\hat{i}-\hat{j}+\hat{k}$ respectively is

If magnitudes of vectors $\vec{a}, \vec{b}, \vec{c}$ are $3, 4,$ and $5$ respectively,and $\vec{a}$ is perpendicular to $\vec{b} + \vec{c}$,$\vec{b}$ is perpendicular to $\vec{c} + \vec{a}$,and $\vec{c}$ is perpendicular to $\vec{a} + \vec{b}$,then find the value of $|\vec{a} + \vec{b} + \vec{c}|$.