If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=2$ and $|\vec{b}|=3$, then the maximum value of $3|(3 \vec{a}+2 \vec{b})|+4|(3 \vec{a}-2 \vec{b})|$ is :

Let P be a point in the plane of the vectors $\overrightarrow{AB}=3\hat{i} + \hat{j} - \hat{k}$ and $\overrightarrow{AC}=\hat{i} - \hat{j} + 3\hat{k}$ such that P is equidistant from the lines AB and AC. If $|\overrightarrow{AP}| = \frac{\sqrt{5}}{2}$, then the area of the triangle ABP is :

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying

$$ |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9 \text { and }|2 \vec{a}+k \vec{b}+k \vec{c}|=3 \text {, } $$

the positive value of k is :

Let $\vec{a}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{c}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to