NIMCET 2016 Question Paper with solutions

Find two vectors in the plane.
We will do this by finding the vector from $\left(1,-1,0\right)\left(2,1,-2\right)$ and from $\left(1,-1,0\right)\left(-1,1,2\right).$
As all three points are in the plane, so will each of those vectors.
$\overrightarrow{{\mathit{u}}_1}=\left(1,-1,0\right)-\left(2,1,-2\right)=\left(-1,-2,2\right)$
$\overrightarrow{{\mathit{u}}_2}=\left(1,-1,0\right)-\left(-1,1,2\right)=\left(2,-2,2\right)$
If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces vector perpendicular to both, we will use the cross product of $\overrightarrow{{\mathit{u}}_1}$ and $\overrightarrow{{\mathit{u}}_2}$ to find a vector $\overrightarrow{\mathit{u}}$ perpendicular to the plane containing them.
$\Rightarrow \overrightarrow{\mathit{u}}=\overrightarrow{{\mathit{u}}_1}\times \overrightarrow{{\mathit{u}}_2}$
$\Rightarrow \overrightarrow{\mathit{u}}=|\begin{array}{ccc}\overrightarrow{\mathit{i}}& \overrightarrow{\mathit{j}}& \overrightarrow{\mathit{k}}\\ -1& -2& 2\\ 2& -2& 2\end{array}|$
$\Rightarrow \overrightarrow{\mathit{u}}=\overrightarrow{\mathit{i}(}-4+4)-\overrightarrow{\mathit{j}(}-2-4)+\overrightarrow{\mathit{k}}\left(2+4\right)$
$\Rightarrow \overrightarrow{\mathit{u}}=6\overrightarrow{\mathit{j}}+6\overrightarrow{\mathit{k}}$
Hence, the vector perpendicular to the plane passing through $\left(1,-1,0\right)\left(2,1,-2\right)$ and $\left(-1,1,2\right)$ is $\overrightarrow{\mathit{u}}=6\overrightarrow{\mathit{j}}+6\overrightarrow{\mathit{k}}$