Explanation

(b)
$\because\ {\overrightarrow{\mathbf{F}}}_{1} = \frac{5(6\widehat{\mathbf{i}} + 2\widehat{\mathbf{j}} + 3\widehat{\mathbf{k}})\ }{7},{\overrightarrow{\mathbf{F}}}_{2} = \ \frac{3(3\widehat{\mathbf{i}} - 2\widehat{\mathbf{j}} + 6\widehat{\mathbf{k}})\ }{7}$
${\overrightarrow{\mathbf{F}}}_{3} = \frac{1(2\widehat{\mathbf{i}} - 3\widehat{\mathbf{j}} - 6\widehat{\mathbf{k}})\ }{7}$
And $\overrightarrow{\mathbf{F}} = {\overrightarrow{\mathbf{F}}}_{1} + {\overrightarrow{\mathbf{F}}}_{2} + {\overrightarrow{\mathbf{F}}}_{3}$
$= \frac{1}{7}(30\widehat{\mathbf{i}} + 10\widehat{\mathbf{j}} + 15\widehat{\mathbf{k}} + 9\widehat{\mathbf{i}} - 6\widehat{\mathbf{j}} + 18\widehat{\mathbf{k}} + 2\widehat{\mathbf{i}} - 3\widehat{\mathbf{j}} - 6\widehat{\mathbf{k}})$
$= \frac{1}{7}(\ 41\widehat{\mathbf{i}} + \widehat{\mathbf{j}} + 27\widehat{\mathbf{k}})$
and $\overrightarrow{\mathbf{AB}} = 5\widehat{\mathbf{i}} - \widehat{\mathbf{j}} + \widehat{\mathbf{k}} - 2\widehat{\mathbf{i}} + \widehat{\mathbf{j}} + 3\widehat{\mathbf{k}}$
$= 3\widehat{i} + 4\widehat{\mathbf{k}}$
$\therefore$ Work done $= \frac{1}{7}\ \left\lbrack 41\widehat{\mathbf{i}} + \widehat{\mathbf{j}} + 27\widehat{\mathbf{k}}\ \right\rbrack \bullet \left\lbrack \ 3\widehat{\mathbf{i}} + 4\widehat{\mathbf{k}}\ \right\rbrack$
$= \frac{1}{7}\ \lbrack\ 123 + 108\ \rbrack = 33$ unit