Find the equation of the normal to the circle 3x2 + 3y2 + 5x − 3y − 22 = 0 at the point (2,1) and the length of the diameter.

The equation of the circle is $3x^2 + 3y^2 + 5x - 3y - 22 = 0$.
Divide by 3 to get the standard form $x^2 + y^2 + \frac{5}{3}x - y - \frac{22}{3} = 0$.
Comparing this to the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$, we have:
$2g = \frac{5}{3} \implies g = \frac{5}{6}$
$2f = -1 \implies f = -\frac{1}{2}$
The center of the circle is $(-g, -f) = \left(-\frac{5}{6}, \frac{1}{2}\right)$.

The normal to a circle at any point passes through the center of the circle. We need to find the equation of the line passing through the given point $(2,1)$ and the center $\left(-\frac{5}{6}, \frac{1}{2}\right)$.
First, calculate the slope $m$ of the normal:

Now, use the point-slope form $y - y_1 = m(x - x_1)$ with the point $(2,1)$:

The equation of the normal is $\boxed{\mathbf{3x - 17y + 11 = 0}}$.

The normal cuts the circle at two points, $(2,1)$ and $\left(-\frac{11}{3}, 0\right)$. Since the normal passes through the center, these two points are the endpoints of a diameter.
Use the distance formula to find the length of the diameter $D$:

The length of the diameter is $\boxed{\mathbf{\frac{\sqrt{298}}{3}}}$.