NIMCET 2019 Question Paper with solutions

Let F be the force
Then $\mathit{F}=\frac{78\left(2\stackrel{\bigwedge }{\mathit{i}}+2\stackrel{\bigwedge }{\mathit{j}}+\stackrel{\bigwedge }{\mathit{k}}\right)}{\sqrt{4+4+1}}=\frac{78}{3}\left(2\stackrel{\bigwedge }{\mathit{i}}+2\stackrel{\bigwedge }{\mathit{j}}+\stackrel{\bigwedge }{\mathit{k}}\right)$
$\mathit{F}=26\left(2\stackrel{\bigwedge }{\mathit{i}}+2\stackrel{\bigwedge }{\mathit{j}}+\stackrel{\bigwedge }{\mathit{k}}\right)$
Force acts at point P(2, 3, 5) the moment of F acting at P about a line in the direction $\left(12\stackrel{\bigwedge }{\mathit{i}}+3\stackrel{\bigwedge }{\mathit{j}}+4\stackrel{\bigwedge }{\mathit{k}}\right)$ is equal to resolve part along the line of moment of F
$\mathit{O}\mathit{P}=2\stackrel{\bigwedge }{\mathit{i}}+3\stackrel{\bigwedge }{\mathit{j}}+5\stackrel{\bigwedge }{\mathit{k}}$
$\mathit{M}=\mathit{O}\mathit{P}\times \mathit{F}=26|\begin{array}{ccc}\stackrel{\bigwedge }{\mathit{i}}& \stackrel{\bigwedge }{\mathit{j}}& \stackrel{\bigwedge }{\mathit{k}}\\ 2& 3& 5\\ 2& 2& 1\end{array}|$
$=26\left(-7\stackrel{\bigwedge }{\mathit{i}}+8\stackrel{\bigwedge }{\mathit{j}}-2\stackrel{\bigwedge }{\mathit{k}}\right)$
Let α be unit vector in the direction $12\stackrel{\bigwedge }{\mathit{i}}+3\stackrel{\bigwedge }{\mathit{j}}+4\stackrel{\bigwedge }{\mathit{k}}$
$\stackrel{\bigwedge }{\mathit{a}}=\frac{12\stackrel{\bigwedge }{\mathit{i}}+3\stackrel{\bigwedge }{\mathit{j}}+4\stackrel{\bigwedge }{\mathit{k}}}{\left|12\stackrel{\bigwedge }{\mathit{i}}+3\stackrel{\bigwedge }{\mathit{j}}+4\stackrel{\bigwedge }{\mathit{k}}\right|}=\frac{1}{13}\left(12\stackrel{\bigwedge }{\mathit{i}}+3\stackrel{\bigwedge }{\mathit{j}}+4\stackrel{\bigwedge }{\mathit{k}}\right)$
The moment of F about the given line is
$=2\left(-84+24-8\right)=-136$