The points $Q(1, 3)$ and $R(7, 0)$ lie on the line $l_1$, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 11 - 2008 - Paper 1

The gradient of line $QR$ is calculated as follows:

$\text{Gradient of } QR = \frac{0 - 3}{7 - 1} = \frac{-3}{6} = -\frac{1}{2}$

Since line $l_2$ is perpendicular to $l_1$, its gradient will be the negative reciprocal:

$\text{Gradient of } l_2 = 2$

Using point $Q(1, 3)$, we can use the point-slope form to write the equation:

$y - 3 = 2(x - 1)$

Rearranging gives us:

$y = 2x + 1$

To find the coordinates of point $P$, we need to determine where line $l_2$ intersects the $y$-axis. This occurs when $x = 0$:

$P = (0, 2(0) + 1) = (0, 1)$

Thus, the coordinates of $P$ are $P(0, 1)$.

To calculate the area of triangle $\triangle PQR$, we can use the formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Here, we can take the base as the distance $QR$ which is $3\sqrt{5}$, and the height is the perpendicular distance from point $P(0, 1)$ to line $QR$.

The coordinates of $Q$ and $R$ are $Q(1, 3)$ and $R(7, 0)$. We can find the area using the determinant method:

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substituting $P(0, 1)$, $Q(1, 3)$, and $R(7, 0)$ into the formula:

$\text{Area} = \frac{1}{2} \left| 0(3 - 0) + 1(0 - 1) + 7(1 - 3) \right|$

Calculating this gives:

$\text{Area} = \frac{1}{2} \left| 0 - 1 - 14 \right| = \frac{1}{2} \left| -15 \right| = 7.5$

Thus, the area of triangle $\triangle PQR$ is $7.5$.