Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$ is equal to

Let $\vec{a} = 3 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 2 \hat{k}$. The projection of the sum of the vectors $\vec{a}$ and $\vec{b}$ on the vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is:

Let $\overrightarrow{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=((\overrightarrow{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

The locus of the point $P(\vec{r})$ which forms a triangle $ABP$ of area $1$ sq. unit with the fixed points $A(\hat{i})$ and $B(\hat{j})$ is

If $|\overline{a}|=3$,$|\overline{b}|=5$,$\overline{b} \cdot \overline{c}=10$,the angle between $\overline{b}$ and $\overline{c}$ is $\frac{\pi}{3}$,and $\overline{a}$ is perpendicular to $\overline{b} \times \overline{c}$,then the value of $|\overline{a} \times(\overline{b} \times \overline{c})|$ is

The acute angle $\theta$ between the vector $\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ and the plane containing the vectors $\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$ is: