If $a, b, c$ are three linearly independent vectors and there exists a non-zero scalar triad $(l, m, n)$ such that $l(3a + 2b + c) + m(2a + 2b + 3c) + n(a + 2b + 5c) = 0$, then:

Let the three sides of a triangle $ABC$ be represented by the vectors $\vec{AB} = 2\hat{i}-\hat{j}+\hat{k}$,$\vec{BC} = 3\hat{i}-4\hat{j}-4\hat{k}$,and $\vec{CA} = \hat{i}-3\hat{j}-5\hat{k}$. Let $G$ be the centroid of the triangle $ABC$. Then $6(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2)$ is equal to

If $\vec{r} = 3\hat{i} + 2\hat{j} - 5\hat{k}$,$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$,and $\vec{c} = -2\hat{i} + \hat{j} - 3\hat{k}$ such that $\vec{r} = \lambda\vec{a} + \mu\vec{b} + \gamma\vec{c}$,then -

The position vector of a point at a distance of $3\sqrt{11}$ units from $i - j + 2k$ on a line passing through the points $i - j + 2k$ and $3i + j + k$ is

If $D, E, F$ are respectively the midpoints of $AB, AC$ and $BC$ in $\Delta ABC$,then $\overrightarrow{BE} + \overrightarrow{AF} = $

If $\vec{a}=\vec{b}+\vec{c},$ then is it true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$ ? Justify your answer.