1. Key idea: conserve internal energy

For a perfect (ideal) gas, the internal energy is proportional to the number of molecules and the temperature:

$U = \text{constant} \times nT$

For the same kind of “perfect gas treatment” used in such questions, we take the internal energy per molecule proportional to $T$. Hence for each gas,

$U_i \propto n_i T_i$

When the gases are mixed with no loss of energy, total internal energy is conserved.

2. Initial total internal energy

$U_{\text{initial}} \propto n_1T_1 + n_2T_2 + n_3T_3$

3. Final state after mixing

Let the final equilibrium temperature be $T$.

Total number of molecules after mixing:

$n_1+n_2+n_3$

So the final internal energy is

$U_{\text{final}} \propto (n_1+n_2+n_3)T$

4. Apply conservation of energy

$n_1T_1 + n_2T_2 + n_3T_3 = (n_1+n_2+n_3)T$

Therefore,

$T = \frac{n_1T_1+n_2T_2+n_3T_3}{n_1+n_2+n_3}$

5. Check dependence on molecular masses

The molecular masses $m_1,m_2,m_3$ do not appear in the final result, because for an ideal gas the average internal energy depends on temperature (and amount of gas), not directly on molecular mass.

6. Compare with options

Option A:

$\frac{n_1T_1+n_2T_2+n_3T_3}{n_1+n_2+n_3}$

This matches exactly.

So, the correct answer is A.