The image positions are calculated lens by lens using the lens formula:
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies v = \frac{uf}{u + f}.\]
For lens 1 (\(f_1 = 10 \, \text{cm}, u = -30 \, \text{cm}\)):
\[v = \frac{(-30) \times 10}{-30 + 10} = \frac{-300}{-20} = 15 \, \text{cm}.\]
The image forms 15 cm to the right of the first lens.
For lens 2 (\(f_2 = -10 \, \text{cm}, u = 10 \, \text{cm}\)):
\[v = \frac{10 \times (-10)}{10 - 10} = \infty.\]
The image is formed at infinity due to parallel rays after the second lens.
For lens 3 (\(f_3 = 30 \, \text{cm}, u = -\infty\)):
\[v = f_3 = 30 \, \text{cm}.\]
Thus, the final image is formed 30 cm to the right of the third lens.
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: