Explanation

(a) The given condition will be satisfied only if one source $(S_{1})$ placed on one side such that $u < f$ ($i.e.$, it lies under the focus). The other source $(S_{2})$ is placed on the other side of the lens such that $u > f\ (i.e.$, it lies beyond the focus)
It $S_{1}$ is the object for lens then $\frac{1}{f} = \frac{1}{- y} - \frac{1}{- x}$
$\Rightarrow \frac{1}{y} = \frac{1}{x} - \frac{1}{f}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
If $S_{2}$ is the object for lens then
$\frac{1}{f} = \frac{1}{+ y} - \frac{1}{- (24 - x)} \Rightarrow \frac{1}{y} = \frac{1}{f} - \frac{1}{(24 - x)}\ \ldots(ii)$
From equations (i) and (ii)
$\frac{1}{x} - \frac{1}{f} = \frac{1}{f} - \frac{1}{(24 - x)} \Rightarrow \frac{1}{x} + \frac{1}{(24 - x)} = \frac{2}{f} = \frac{2}{9}$
$\Rightarrow x^{2} - 24x + 108 = 0$. After solving the equation $x = 18\ cm,\ 6\ cm$