Step 1: Set up the distance from a point on the curve to the given line.
A general point on $y = x^2+4x+3$ is $P = (t,\, t^2+4t+3)$. The distance from $P$ to the line $3x - y + 2 = 0$ is $D(t) = \frac{|3t - (t^2+4t+3) + 2|}{\sqrt{9+1}} = \frac{|-t^2 - t - 1|}{\sqrt{10}}$.
Step 2: Simplify the expression inside the absolute value.
$-t^2 - t - 1 = -(t^2+t+1)$. The discriminant of $t^2+t+1$ is $1-4 = -3 < 0$, so $t^2+t+1 > 0$ for all real $t$. Hence $D(t) = \frac{t^2+t+1}{\sqrt{10}}$.
Step 3: Minimize $D(t)$ by minimizing $h(t) = t^2 + t + 1$.
$h'(t) = 2t + 1 = 0 \Rightarrow t = -\frac{1}{2}$. Since $h''(t) = 2 > 0$, this is a minimum.
Step 4: Find the $y$-coordinate at $t = -1/2$.
$y = \left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) + 3 = \frac{1}{4} - 2 + 3 = \frac{1}{4} + 1 = \frac{5}{4}$.
Step 5: Verify the point is on the curve.
At $t = -1/2$: $y = 1/4 - 2 + 3 = 5/4$. Yes, the point $(-1/2,\, 5/4)$ lies on the curve $y = x^2+4x+3$.
Step 6: State the final answer.
The point on the curve closest to the line is: \[\boxed{\left(-\dfrac{1}{2},\, \dfrac{5}{4}\right)}\]