The dimensions of Planck's constant is $$h = [{M^1}{L^3}{T^{ - 2}}]$$

Speed of light is $$c = [{L^1}{T^{ - 1}}]$$

Gravitational constant is $$G = [{M^{ - 1}}{L^3}{T^{ - 2}}]$$

Let $$L \propto {h^x}{c^y}{G^z}$$.

$$[L] = {[{M^1}{L^2}{T^{ - 1}}]^x}{[{L^1}{T^{ - 1}}]^y}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^z}$$

Comparing, we get

$$\left. {\matrix{ \hfill {x - z = 0} \cr \hfill {2x + y + 3z = 1} \cr \hfill { - x - y - 2z = 0} \cr } } \right\}$$

Solving, we get

$$x = z$$

$$y + 5x = 1$$

$$ - y - 3x = 0$$

or, $$2x = 1$$

$$x = {1 \over 2} = z$$

$$y = - {3 \over 2}$$

Therefore,

$$L \propto {h^{1/2}}{C^{ - 3/2}}{G^{1/2}}$$

$$L \propto \sqrt h $$

$$L \propto \sqrt G $$

Let $$M \propto {h^a}{C^b}{G^c}$$

$$[M] = {[{M^1}{L^2}{T^{ - 1}}]^a}{[{L^1}{T^{ - 1}}]^b}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^c}$$

Comparing, we get $$a - c = 1$$.

$$\left. \matrix{ 2a + b + 3c = 0 \hfill \cr - a - b - 2c = 0 \hfill \cr} \right\}a + c = 0$$

Therefore,

$$2a = 1$$

$$a = {1 \over 2}$$

$$c = - {1 \over 2}$$

$$b = {1 \over 2}$$

Therefore,

$$M \propto {h^{1/2}}{c^{1/2}}{G^{ - 1/2}}$$

$$M \propto \sqrt c $$