1. Given data

• Wavelength: $\lambda = 1\,\text{m}$
• Sources $S_1$ and $S_2$ are in phase.
• Separation between sources: $S_1S_2 = 1.5\,\text{m}$
• Listener is initially directly in front of $S_2$ and at a distance $2\,\text{m}$ from $S_2$.
• At this position, intensity is minimum.

We need the distance $d$ from $S_1$ when the adjacent maximum is observed, while the listener keeps distance from $S_2$ fixed.

2. Find the initial path difference

Since the listener is directly in front of $S_2$, his distance from $S_2$ is $LS_2 = 2\,\text{m}.$

Using the geometry in the figure, $S_1$ is $1.5\,\text{m}$ away horizontally from $S_2$, so $LS_1 = \sqrt{(2)^2 + (1.5)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5\,\text{m}.$

Hence, initial path difference is $\Delta = LS_1 - LS_2 = 2.5 - 2 = 0.5\,\text{m} = \frac{\lambda}{2}.$

This is consistent with minimum intensity, because for in-phase sources, minima occur when $\Delta = \left(n + \frac12\right)\lambda.$

3. Condition for adjacent maximum

The listener moves away from $S_1$ while keeping distance from $S_2$ fixed at $2\,\text{m}$.

So $LS_2$ remains $2\,\text{m}$, while $LS_1 = d$ changes.

Path difference becomes $\Delta = d - 2.$

Initially, minimum occurs at $\Delta = \frac{\lambda}{2} = 0.5\,\text{m}.$

The adjacent maximum will occur at the nearest constructive interference condition: $\Delta = 1\lambda = 1\,\text{m}.$

So, $d - 2 = 1$ $d = 3\,\text{m}.$

4. Check options

• A: $12\,\text{m}$ $\times$
• B: $2\,\text{m}$ $\times$
• C: $3\,\text{m}$ $\checkmark$
• D: $5\,\text{m}$ $\times$

Thus, the correct option is $\boxed{3\,\text{m}}.$

So, Option C is correct.