$A$ unit vector perpendicular to the plane determined by the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ is:

Let $ABC$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{AB}=\hat{i}+2 \hat{j}-7 \hat{k}$,$\overrightarrow{BC}=a \hat{i}+b \hat{j}+c \hat{k}$ and $\overrightarrow{AC}=6 \hat{i}+d \hat{j}-2 \hat{k}$,where $d>0$. Then the square of the length of the largest side of the triangle $ABC$ is:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$ are given vectors, then a vector $\vec{b}$ satisfying the equations $\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=3$ is

The adjacent sides of a parallelogram are $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. Find its area.

Let $A(2, 3, 0)$, $B(0, 3, 2)$ and $C(4, 0, 3)$ be the vertices of a triangle. Find the area of the triangle.

Find the unit vector which is perpendicular to the vector $5i + 2j + 6k$ and coplanar with the vectors $2i + j + k$ and $i - j + k$.