Time ($$T$$), velocity ($$C$$) and angular momentum ($$h$$) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

We are told that instead of the usual base quantities mass ($$M$$), length ($$L$$) and time ($$T$$), a new set is being adopted: time $$T$$ (which keeps its old meaning), velocity $$C$$ and angular momentum $$h$$. Our task is to write the dimension of mass in terms of these three new fundamental quantities.

Let us assume that mass can be written as a product of powers of the new base quantities:

$$M = T^{\,a}\;C^{\,b}\;h^{\,c}$$

Here $$a, b, c$$ are the numerical exponents we need to determine. To do that we must translate every factor back into the ordinary $$M,\,L,\,T$$ system and then equate exponents.

First we recall or state the dimensional formulae of the new bases in the old system:

1. Time $$T$$ is, of course, just $$[T^1]$$.

2. Velocity $$C$$ has the ordinary dimensions of length per time, so $$C \;\Longrightarrow\; [L^1\,T^{-1}]$$.

3. Angular momentum $$h$$ has the standard dimensional form “momentum × distance’’ i.e. $$[M^1\,L^2\,T^{-1}]$$.

Now substitute these into the assumed product:

$$T^{\,a}\;C^{\,b}\;h^{\,c} \;=\; \bigl[T^1\bigr]^{\,a}\; \bigl[L^1\,T^{-1}\bigr]^{\,b}\; \bigl[M^1\,L^2\,T^{-1}\bigr]^{\,c}$$

Collecting the like bases term-by-term gives

$$ = M^{\,c}\; L^{\,b + 2c}\; T^{\,a - b - c}. $$

We want this overall dimension to equal the single pure mass dimension $$[M^1]$$, which is $$M^1\,L^0\,T^0$$. Hence we equate the exponents of $$M$$, $$L$$ and $$T$$ separately:

For $$M$$: $$c \;=\; 1$$.

For $$L$$: $$b + 2c \;=\; 0$$.

For $$T$$: $$a - b - c \;=\; 0$$.

Substituting the first result $$c = 1$$ into the second equation gives

$$b + 2(1) = 0 \;\Longrightarrow\; b = -2.$$

Now substitute $$b = -2$$ and $$c = 1$$ into the third equation:

$$a - (-2) - 1 = 0 \;\Longrightarrow\; a + 2 - 1 = 0 \;\Longrightarrow\; a = -1.$$

Thus the required exponents are $$a = -1,\; b = -2,\; c = 1$$. Putting them back, the dimensional formula for mass in the new base set is

$$[T^{-1}\,C^{-2}\,h^{\,1}] = [T^{-1}C^{-2}h].$$

Comparing with the given options, this matches option A.

Hence, the correct answer is Option A.