Explanation

(d) In Ist case, when charge $\mathcal{+ Q}$ is situated at C.
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Electric potential energy of system
$U_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{(q)( - q)}{2L} + \frac{1}{4\pi\varepsilon_{0}}\frac{( - q)\mathcal{Q}}{L} + \frac{1}{4\pi\varepsilon_{0}}.\frac{q\mathcal{Q}}{L}$
In IInd case, when charge $+ \mathcal{Q}$is moved from C to D.
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Electric potential energy of system in that case
$U_{2} = \frac{1}{4\pi\varepsilon_{0}}.\frac{(q)( - q)}{2L} + \frac{1}{4\pi\varepsilon_{0}}.\frac{q\mathcal{Q}}{3L} + \frac{1}{4\pi\varepsilon_{0}}\frac{( - q)\left( \mathcal{Q} \right)}{L}$
$\therefore Work\ done = \mathrm{\Delta}U = U_{2} - U_{1}$
$= \left( \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{2L} + \frac{1}{4\pi\varepsilon_{0}}\frac{q\mathcal{Q}}{3L} - \frac{1}{4\pi\varepsilon_{0}}\frac{q\mathcal{Q}}{L} \right)$
$- \left( \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{2L} + \frac{1}{4\pi\varepsilon_{0}}\frac{q\mathcal{Q}}{L} + \frac{1}{4\pi\varepsilon_{0}}\frac{q\mathcal{Q}}{L} \right)$
$= \frac{q\mathcal{Q}}{4\pi\varepsilon_{0}}.\left\lbrack \frac{1}{3L} - \frac{1}{L} \right\rbrack$
$= \frac{q\mathcal{Q}}{4\pi\varepsilon_{0}}\frac{(1 - 3)}{3L}$
$= \frac{- 2q\mathcal{Q}}{12\pi\varepsilon_{0}L} = - \frac{q\mathcal{Q}}{6\pi\varepsilon_{0}L}$