Step 1: Name the unknown.
Let the lens sit at distance \(a\) from source \(S_1\); the remaining distance to \(S_2\) is \(b = 24 - a\). The images formed by the two beams passing through the lens are required to land on the very same point.
Step 2: Apply the lens formula to each beam.
Taking the standard convention, the image distance of \(S_1\) is \(v_1\) with \(\dfrac{1}{v_1}=\dfrac{1}{9}-\dfrac{1}{a}\), and the image distance of \(S_2\) is \(v_2\) with \(\dfrac{1}{v_2}=\dfrac{1}{9}-\dfrac{1}{b}\).
Step 3: Coincidence condition by distances.
Since the beams enter from opposite sides, their images meet at one point when the image of the first lies on the same physical spot as the image of the second, giving \(v_1 = -v_2\). Substituting \(b = 24 - a\) and simplifying the equation \(\dfrac{9a}{a-9} + \dfrac{9(24-a)}{15-a}=0\) leads to \(a^2 - 24a + 108 = 0\).
Step 4: Factor the quadratic.
\(a^2 - 24a + 108 = (a-6)(a-18) = 0\), so \(a = 6\) cm or \(a = 18\) cm. These are the same setup seen from the two sources.
Step 5: Verify with the 18 cm placement.
Take \(a = 18\) cm: image of \(S_1\) is \(v_1 = \dfrac{9\times18}{18-9}=18\) cm (real, on far side), and for \(S_2\) at \(b = 6\) cm, \(v_2 = \dfrac{9\times6}{6-9}=-18\) cm (virtual, same far side), so once again both images meet 18 cm from the lens. The placement is therefore 6 cm from one source and 18 cm from the other.
\[\boxed{a = 6\ \text{cm or } 18\ \text{cm from a source}}\]