1. Write dimensions of the given fundamental constants

We express everything in terms of $M, L, T$.

• Velocity of light: $[c] = LT^{-1}$

• Gravitational constant: From Newton's law, $F = \frac{Gm_1m_2}{r^2}$ so $[G] = \frac{[F][r^2]}{[M]^2} = \frac{(MLT^{-2})L^2}{M^2} = M^{-1}L^3T^{-2}$

• Planck's constant: $[h] = [\text{energy}]\cdot [\text{time}] = (ML^2T^{-2})T = ML^2T^{-1}$

2. Assume mass can be written in terms of $h, c, G$

Let $[M] = [h]^a[c]^b[G]^d$

Substitute dimensions: $[M] = (ML^2T^{-1})^a(LT^{-1})^b(M^{-1}L^3T^{-2})^d$

So, $[M] = M^{a-d}L^{2a+b+3d}T^{-a-b-2d}$

Since the left side is just mass, $[M] = M^1L^0T^0$

Therefore, equate powers:

• For $M$: $a-d = 1$
• For $L$: $2a+b+3d = 0$
• For $T$: $-a-b-2d = 0$

3. Solve the equations

From the third equation: $a+b+2d=0 \implies b=-a-2d$

Substitute into the second equation: $2a+(-a-2d)+3d=0$ $a+d=0 \implies a=-d$

Now use the first equation: $a-d=1$ $(-d)-d=1$ $-2d=1$ $d=-\frac{1}{2}$

Hence, $a=\frac{1}{2}$

And $b=-a-2d=-\frac{1}{2}-2\left(-\frac{1}{2}\right)=\frac{1}{2}$

4. Final expression

Thus, $[M] = [h^{1/2}c^{1/2}G^{-1/2}]$

So the correct option is: $\boxed{\text{C}}$