If the area of the triangle with vertices $\hat{i}+y \hat{j}$,$\hat{i}+2 \hat{k}$,and $3 \hat{j}+\hat{k}$ is $\sqrt{6}$ sq. units,then the values of $y$ are

Consider $P(1, 2, -3)$,$Q(-2, 1, -4)$,and $R(3, 4, -2)$. Let $\vec{B} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$,where $A_x, A_y$,and $A_z$ are the projections of the area of triangle $PQR$ on the $yz, zx$,and $xy$ planes,respectively. Then,the value of $|\vec{B}|^2$ is:

If $a = i + 2j - 2k$,$b = 2i - j + k$ and $c = i + 3j - k$,then $a \times (b \times c)$ is equal to

Find a unit vector perpendicular to the vector $2\hat{i} - \hat{j} + 2\hat{k}$ and coplanar with the vectors $\hat{i} + 2\hat{j} - \hat{k}$ and $2\hat{i} + \hat{j} - \hat{k}$.

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. Let $\vec c$ be a vector such that $|\vec c - \vec a| = 3$,$|(\vec a \times \vec b) \times \vec c| = 3$,and the angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$. Then $\vec a \cdot \vec c$ is equal to:

Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}$ be a vector such that $|\vec{c}-\vec{a}|=3$. If $\vec{p}=\vec{a} \times \vec{b}$,then the angle between $\vec{p}$ and $\vec{c}$ is $\frac{\pi}{6}$ and $|\vec{p} \times \vec{c}|=3$. Thus,$\vec{a} \cdot \vec{c}$ is equal to: