If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = -3(\hat{i} - \hat{k})$. Then the triplet $(\alpha, \beta, \gamma)$ is

If $\overrightarrow{AB} = 2\hat{i} + 3\hat{j} - 6\hat{k}$ and $\overrightarrow{BC} = 6\hat{i} - 2\hat{j} + 3\hat{k}$ are the vectors along two sides of a triangle $ABC$,then the perimeter of triangle $ABC$ is:

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and the maximum value of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ is $k$,then $k(2|\vec{a}|^2+3|\vec{b}|^2-4|\vec{c}|^2) = $

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors and $|\overrightarrow{a}+\overrightarrow{b}|=1$,then $|\overrightarrow{a}-\overrightarrow{b}|$ is equal to

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 the vectors $2 \hat{i}-3 \hat{j}+6 \hat{k}$ and $\vec{b}$ are collinear and $|\vec{b}|=14$,then $\vec{b}$ has the value