Inspired by Ashish Siva

Let $\sqrt{3+\sqrt{3}+\sqrt{6}+\sqrt{2}} = \sqrt{(x+y\sqrt{2}+z\sqrt{3})^2} = x+y\sqrt{2}+z\sqrt{3}$ .

Squaring both sides, we have:

$\begin{aligned} (x+y\sqrt{2}+z\sqrt{3})^2 & = 3+\sqrt{3}+\sqrt{6}+\sqrt{2} \\ x^2 + 2y^2 + 3z^2 + 2xy\sqrt{2} + 2yz\sqrt{6} + 2zx\sqrt{3} & = 3+\sqrt{3}+\sqrt{6}+\sqrt{2} \end{aligned}$

Equating the coefficients, we have:

$\begin{cases} 2xy = 1 &...(1) \\ 2yz = 1 & ...(2) \\ 2zx = 1 &...(3) \end{cases}$

From $\dfrac{(1)}{(2)}: \quad \dfrac{x}{z} = 1 \quad \Rightarrow x = z$ .

Similarly, from $\dfrac{(2)}{(3)}: \quad \Rightarrow x = y$ .

From $(1): \quad 2xy = 1 \quad \Rightarrow 2x^2 = 1 \quad \Rightarrow x = y = z = \dfrac{1}{\sqrt{2}}$

Therefore,

$\begin{aligned} \sqrt{3+\sqrt{3}+\sqrt{6}+\sqrt{2}} & = x+y\sqrt{2}+z\sqrt{3} \\ & = \sqrt{\frac{1}{2}} + 1 + \sqrt{\frac{3}{2}} \end{aligned}$

$\Rightarrow a + b + c + d + e = 1 + 2 + 3 + 2 + 1 = \boxed{9}$

Add the radical fractions then square it and the original equation. Then just equate the corresponding parts.