Find the co-ordinates of the orthocentre of the triangle formed by the lines
$L_1: y - x = 2$
$L_2: y + 2x = 8$
$L_3: 3y - x = 18$

The vertices of the triangle are the points of intersection of the given lines.

Solving $L_1$ and $L_2$: $y - x = 2$ and $y + 2x = 8 \Rightarrow x = 2, y = 4$. Let $A \equiv (2, 4)$.

Solving $L_1$ and $L_3$: $y - x = 2$ and $3y - x = 18 \Rightarrow x = 6, y = 8$. Let $C \equiv (6, 8)$.

The slope of $L_3$ (line $BC$) is $\dfrac{1}{3}$. The altitude from $A$ to $BC$ is perpendicular to $L_3$, so its slope is $-3$.

Equation of the altitude from $A(2, 4)$ is:
$y - 4 = -3(x - 2) \Rightarrow 3x + y = 10$

The slope of $L_2$ (line $AB$) is $-2$. The altitude from $C$ to $AB$ is perpendicular to $L_2$, so its slope is $\dfrac{1}{2}$.

Equation of the altitude from $C(6, 8)$ is:
$y - 8 = \dfrac{1}{2}(x - 6) \Rightarrow x - 2y = -10$

The orthocentre is the point of intersection of the altitudes. Solving $3x + y = 10$ and $x - 2y = -10$:

Multiplying the first equation by $2$ and adding to the second gives:
$7x = 10 \Rightarrow x = \dfrac{10}{7}$

Substituting $x$ in the first equation gives:
$y = 10 - 3\left(\dfrac{10}{7}\right) = \dfrac{40}{7}$

The orthocentre is $\left(\dfrac{10}{7}, \dfrac{40}{7}\right)$.

Answer: $\left(\dfrac{10}{7}, \dfrac{40}{7}\right)$