The area bounded by the curves $y^2 = 12x$ and $x^2 = 12y$ is divided by the line $x = 3$ in two parts. The area (in square units) of the larger part, is

$\frac{147}{4}$

Let $A_1$ and $A_2$ be areas of regions OPRO and PQRP respectively. Then,

$A_1=\int\limits_0^3\left(\sqrt{12x}-\frac{x^2}{12}\right)dx$ and $A_2=\int\limits_3^{12}\left(\sqrt{12x}-\frac{x^2}{12}\right)dx$

$⇒A_1=\left[2\sqrt{3}\frac{x^{3/2}}{3/2}-\frac{x^3}{36}\right]_0^3$ and $A_2=\left[2\sqrt{3}\frac{x^{3/2}}{3/2}-\frac{x^3}{36}\right]_3^{12}$

$⇒A_1=\frac{4}{\sqrt{3}}×3\sqrt{3}-\frac{27}{36}$

and, $A_2=\frac{4}{\sqrt{3}}\{(12\sqrt{12}-3\sqrt{3})\}-\frac{1}{36}(12^3-3^3)$

$⇒A_1=12-\frac{3}{4}$ and $A_2=\frac{4}{\sqrt{3}}\{24\sqrt{12}-3\sqrt{3}\}-\frac{189}{4}=\frac{147}{4}$

$⇒A_1=\frac{45}{4}$ and $A_2=\frac{147}{4}$

Hence, required answer is $\frac{147}{4}$