To solve this definite integral, we first find the indefinite integral:
\[
\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C.
\]
Here, $a = 3$, so:
\[
\int \sin(3x) \, dx = -\frac{1}{3} \cos(3x) + C.
\]
Now, apply the limits:
\[
\int_{0}^{\pi/6} \sin(3x) \, dx
= \Big[ -\frac{1}{3} \cos(3x) \Big]_{0}^{\pi/6}.
\]
Calculate the cosine values:
\[
\cos\Big(3 \times \frac{\pi}{6}\Big) = \cos\Big(\frac{\pi}{2}\Big) = 0, \cos(0) = 1.
\]
So,
\[
\int_{0}^{\pi/6} \sin(3x) \, dx
= -\frac{1}{3} [\cos(\pi/2) - \cos(0)] = -\frac{1}{3} [0 - 1] = -\frac{1}{3}(-1) = \frac{1}{3}.
\]
Therefore, the answer is $\frac{1{3}$}.