The line $l_1$ has equation $2x - 3y + 12 = 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 1

Since $l_2$ is perpendicular to $l_1$, the gradient of $l_2$ can be found using the negative reciprocal of $\frac{2}{3}$. Therefore,
$m_{l_2} = -\frac{3}{2}.$

To find the coordinates of point B, we set $x=0$ in the original equation $2x - 3y + 12 = 0$:
\begin{align*} 2(0) - 3y + 12 &= 0 \ -3y &= -12 \ y &= 4 \ ext{Thus, B is at } (0,4). \end{align*}

Now we can use point-slope form to find the equation of $l_2$:
$y - y_1 = m(x - x_1)$
Substituting B(0, 4):
$y - 4 = -\frac{3}{2}(x - 0) \ y = -\frac{3}{2}x + 4.$
Thus, the equation of $l_2$ is $y = -\frac{3}{2}x + 4$.

To find the x-intercept of line $l_2$, we set $y = 0$ in the equation:
$0 = -\frac{3}{2}x + 4 \Rightarrow \frac{3}{2}x = 4 \Rightarrow x = \frac{4 \cdot 2}{3} = \frac{8}{3}.$
Thus, point C is at $\left(\frac{8}{3}, 0\right)$.

To find the area of triangle ABC with vertices A(\frac{8}{3}, 0), B(0, 4), and C(0, 0), we can use the formula for the area of a triangle given by vertices:
$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$
Substituting in the coordinates:
$\text{Area} = \frac{1}{2} \left| \frac{8}{3}(4 - 0) + 0(0 - 0) + 0(0 - 4) \right| = \frac{1}{2} \left| \frac{32}{3} \right| = \frac{16}{3}.$
Thus, the area of triangle ABC is $\frac{16}{3}$.