In an interference experiment using two coherent sources of light,...

For destructive interference (minima), the phase difference must be an odd multiple of $\pi$, given by the condition $\Delta\phi = (2n-1)\pi$, where $n = 1, 2, 3, \dots$ represents the order of the minimum.

Given the phase difference $\Delta\phi = 3\pi$, we can equate:
$(2n-1)\pi = 3\pi$
$2n - 1 = 3$
$2n = 4 \implies n = 2$

Therefore, the fringe formed at this point is the second minimum.

Answer: Second minimum