(a) Potential at the mid point of the line joining the charges ${\mathit{Q}}_1$ and $-{\mathit{Q}}_2$ is
$\mathrm{V}=\mathrm{k}\frac{{\mathrm{Q}}_1}{\frac{\mathrm{r}}{2}}+\mathrm{k}\frac{{\mathrm{Q}}_2}{\frac{\mathrm{r}}{2}}$
( $\mathrm{r}$ is the distance between ${\mathrm{Q}}_1$ and $-{\mathrm{Q}}_2$ )
Or, $\mathit{V}=\frac{2\mathit{k}{\mathit{Q}}_1}{\mathit{r}}-\frac{2\mathit{k}{\mathit{Q}}_2}{\mathit{r}}$
Work done to place ${\mathrm{Q}}_3$ is $\mathrm{W}=\mathrm{V}\times {\mathrm{Q}}_3$
Or, $\mathrm{W}=\left(\frac{2\mathit{k}{\mathit{Q}}_1}{\mathit{r}}-\frac{2\mathit{k}{\mathit{Q}}_2}{\mathit{r}}\right)\times {\mathit{Q}}_3$
$\therefore \mathrm{W}=\frac{2\mathit{k}}{\mathit{r}}\left({\mathit{Q}}_1-{\mathit{Q}}_2\right){\mathit{Q}}_3$
(b) Work done will be zero when $\mathrm{V}=0$
Let us consider that a point at a distance $\mathit{x}$ from the charge ${\mathit{Q}}_1$ where the potential will be zero.
So, $\frac{\mathit{k}{\mathit{Q}}_1}{\mathit{x}}-\frac{\mathit{k}{\mathit{Q}}_2}{\mathit{r}-\mathit{x}}=0$
Or, $\frac{{\mathit{Q}}_1}{\mathit{x}}-\frac{{\mathit{Q}}_2}{\mathit{r}-\mathit{x}}=0$
Or, $\frac{{\mathit{Q}}_1}{\mathit{x}}=\frac{{\mathit{Q}}_2}{\mathit{r}-\mathit{x}}$
Or, ${\mathit{Q}}_1\left(\mathit{r}-\mathit{x}\right)={\mathit{Q}}_2\mathit{x}$
$\therefore \mathit{x}=\frac{{\mathit{Q}}_1\mathit{r}}{{\mathit{Q}}_1+{\mathit{Q}}_2}$
So, at a distance $\frac{{\mathit{Q}}_1\mathit{r}}{{\mathit{Q}}_1+{\mathit{Q}}_2}$ from charge ${\mathit{Q}}_1$ on the line joining the two charges, the work done will be zero.