NIMCET 2018 Question Paper with solutions

Let
$\overrightarrow{{\mathit{F}}_1}$ = $4\hat{\mathit{i}}-3\hat{\mathit{j}}+7\hat{\mathit{k}}$ , $\overrightarrow{{\mathit{F}}_2}$ = - $2\hat{\mathit{i}}+2\hat{\mathit{j}}-8\hat{\mathit{k}}$
∴ Resulting force $\overrightarrow{{\mathit{F}}_1}+\overrightarrow{{\mathit{F}}_2}$ = $\left(4\hat{\mathit{i}}-3\hat{\mathit{j}}+7\hat{\mathit{k}}\right)+\left(-2\hat{\mathit{i}}+2\hat{\mathit{j}}-8\hat{\mathit{k}}\right)$
= $2\hat{\mathit{i}}-\hat{\mathit{j}}-\hat{\mathit{k}}$
Since, point is displace from A(5,7,1) to B (2,5,- 6)
∴ $\overrightarrow{\mathit{A}\mathit{B}}$ = (2 - 5) $\hat{\mathit{i}}$ + (5 - 7) $\hat{\mathit{j}}$ + (- 6 - 1) $\hat{\mathit{k}}$
= - 3 $\hat{\mathit{i}}$ - 2 $\hat{\mathit{j}}$ - 7 $\hat{\mathit{k}}$
∴ Work done = $\overrightarrow{\mathit{F}}.\overrightarrow{\mathit{A}\mathit{B}}$
= $\left(2\hat{\mathit{i}}-\hat{\mathit{j}}-\hat{\mathit{k}}\right).\left(-3\hat{\mathit{i}}-2\hat{\mathit{j}}-7\hat{\mathit{k}}\right)$
= - 6 + 2 + 7 = 3 units
No option match