Explanation

(d) For first lens, $\mu_{1} = - 30cm,f_{1} = 10cm$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$or\ \frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$or\ \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$
$or\ v = 15\ cm$
Therefore, image formed by convex lens ($L_{1}$) is at point $I_{1}$ and acts as virtual object for concave lens ($L_{2}$).
The image $I_{1}$ is formed at focus of concave lens (as shown) and so emergent rays will be parallel to the principle axis. For lens $L_{2}\ $, $\mu_{2}$ =15-5=10 cm, $f_{2}$=-10cm. These parallel rays are incident on the third convex lens ($L_{3}$) and will be brought to convergence at the focus of the lens $(L_{3})$
Hence , distance of final image from third lens $L_{3}$
$v_{2} = f_{3} = 30\ cm$