$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\sqrt{6}}{\left(\sqrt{6}\right)^{2}}$$
The square of $\sqrt{6}$ is $6$.$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\sqrt{6}}{6}$$
Use the distributive property to multiply $\sqrt{3}+\sqrt{2}$ by $\sqrt{6}$.$$\frac{\sqrt{3}\sqrt{6}+\sqrt{2}\sqrt{6}}{6}$$
Factor $6=3\times 2$. Rewrite the square root of the product $\sqrt{3\times 2}$ as the product of square roots $\sqrt{3}\sqrt{2}$.$$\frac{\sqrt{3}\sqrt{3}\sqrt{2}+\sqrt{2}\sqrt{6}}{6}$$
Multiply $\sqrt{3}$ and $\sqrt{3}$ to get $3$.$$\frac{3\sqrt{2}+\sqrt{2}\sqrt{6}}{6}$$
Factor $6=2\times 3$. Rewrite the square root of the product $\sqrt{2\times 3}$ as the product of square roots $\sqrt{2}\sqrt{3}$.$$\frac{3\sqrt{2}+\sqrt{2}\sqrt{2}\sqrt{3}}{6}$$
Multiply $\sqrt{2}$ and $\sqrt{2}$ to get $2$.$$\frac{3\sqrt{2}+2\sqrt{3}}{6}$$