A line passes through point P(4, 9) and Q(3, 2) is perpendicular to the line 3y + ax + 3 = 0. Find the value of a and write down the equation of line PQ.

The coordinates of P are $(x_1, y_1) = (4, 9)$ and Q are $(x_2, y_2) = (3, 2)$.
The gradient $m_{PQ}$ is given by:

$$m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 9}{3 - 4} = \frac{-7}{-1} = 7$$

Rearrange the equation into the form $y = mx + c$:

$$3y = -ax - 3$$ $$y = -\frac{a}{3}x - 1$$

The gradient of this line, $m_L$, is $-\frac{a}{3}$.

If two lines are perpendicular, the product of their gradients is -1.

$$m_{PQ} \times m_L = -1$$ $$7 \times \left(-\frac{a}{3}\right) = -1$$ $$-\frac{7a}{3} = -1$$

Multiply both sides by -3:

$$7a = 3$$ $$a = \frac{3}{7}$$

We have the gradient $m_{PQ} = 7$ and can use either point P(4, 9) or Q(3, 2). Using point P(4, 9) and the point-slope form $y - y_1 = m(x - x_1)$:

$$y - 9 = 7(x - 4)$$ $$y - 9 = 7x - 28$$ $$y = 7x - 28 + 9$$ $$y = 7x - 19$$

The equation can also be written as $7x - y - 19 = 0$.

The value of $a$ is $\boxed{\frac{3}{7}}$ and the equation of line PQ is $\boxed{y = 7x - 19}$.