What would be the position of image formed by the lens combination given in the figure below?

6 cm to the right of third lens

The correct answer is Option (1) → 6 cm to the right of third lens

for the first lens ($f_1=+10cm$)

$\frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u_1}$

$\frac{1}{10}=\frac{1}{v_1}-\frac{1}{-20}$

$⇒\frac{1}{v}=\frac{1}{10}+\frac{1}{20}=\frac{3}{20}$

$⇒v=\frac{20}{3}cm$

for the first lens ($f_2=-10cm$)

$\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}$

$\frac{1}{-10}=\frac{1}{v_2}-\frac{3}{10}$

$⇒\frac{1}{v_2}=\frac{-1}{10}+\frac{3}{10}=\frac{-2}{10}$

$⇒v_2=5cm$

for the first lens ($f_3=+10cm$)

$\frac{1}{f_3}=\frac{1}{v_3}-\frac{1}{u_3}$

$\frac{1}{10}=\frac{1}{v_2}-\frac{1}{15}$

$⇒\frac{1}{v_3}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}$

$⇒\frac{1}{v_3}=\frac{5}{30}cm$

$⇒v_3=6cm$