Let ABC be the triangle such that the equations of lines AB and AC be $$3y - x = 2$$ and $$x + y = 2$$, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to

Lines AB: $$3y - x = 2$$ and AC: $$x + y = 2$$, with B and C on the x-axis.

Point B lies on AB and the x-axis ($$y = 0$$): $$-x = 2$$, so $$B = (-2, 0)$$.

Point C lies on AC and the x-axis ($$y = 0$$): $$x = 2$$, so $$C = (2, 0)$$.

Point A is the intersection of AB and AC: substituting $$x = 2 - y$$ into $$3y - x = 2$$:

$$3y - (2-y) = 2$$, giving $$4y = 4$$, so $$y = 1$$ and $$x = 1$$. Thus $$A = (1, 1)$$.

Finding the orthocentre P of triangle ABC:

Since BC lies on the x-axis (horizontal), the altitude from A to BC is vertical: $$x = 1$$.

The slope of AC is $$-1$$, so the altitude from B to AC has slope $$1$$. Through $$B(-2, 0)$$: $$y = x + 2$$.

Intersection of $$x = 1$$ and $$y = x + 2$$: $$P = (1, 3)$$.

Area of triangle PBC with vertices $$P(1,3)$$, $$B(-2,0)$$, $$C(2,0)$$:

Base $$BC = 4$$, height = y-coordinate of P = 3.

Area $$= \frac{1}{2} \times 4 \times 3 = 6$$

Hence, the correct answer is Option D.