In the given problem, a molecule is undergoing two separate, simultaneous first-order reactions. Each of these reactions has its own rate constant and half-life. The half-life ($T$) of a first-order reaction is related to its rate constant ($k$) by the equation:

$$T = \frac{\ln(2)}{k}$$

When two or more first-order reactions are occurring independently and simultaneously (also known as parallel or competing reactions), the overall rate constant for the disappearance of the reactant is the sum of the rate constants for the individual reactions:

$$k_{\text{total}} = k_1 + k_2$$

Conversely, the overall half-life for the disappearance of the reactant is determined by the reciprocal of the sum of the reciprocals of the individual half-lives:

$$\frac{1}{T_{\text{total}}} = \frac{1}{T_1} + \frac{1}{T_2}$$

This equation reflects the fact that the reactant is disappearing more quickly than it would due to any single reaction, because it is being consumed by two reactions at once.

Substituting the given half-lives of $12$ minutes and $3$ minutes into this equation gives:

$$\begin{align} \frac{1}{T_{\text{total}}} &= \frac{1}{12\text{ min}} + \frac{1}{3\text{ min}} \\\\ &= \frac{5}{12\text{ min}} \end{align}$$

Solving for $T_{\text{total}}$ then gives:

$$T_{\text{total}} = \frac{12}{5}\text{ min} = 2.4\text{ min}$$

This is the time taken for $50\%$ of the reactant to be consumed when both reactions are occurring. Rounding to the nearest integer gives an answer of $2$ minutes.