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 the x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to:

Step 1: Find the coordinates of vertices A, B, and C

First, we find the coordinates of the vertices of triangle ABC. Vertex A is the intersection of lines AB ($x - 3y = -2$) and AC ($x + y = 2$). Solving these two equations simultaneously gives $x=1$ and $y=1$, so $A=(1{,}1)$. Since B and C lie on the $x$ -axis, their $y$ -coordinates are 0. For B, substitute $y=0$ into the equation of line AB ($3y - x = 2$) to get $x = -2$, so $B=(-2{,}0)$. For C, substitute $y=0$ into the equation of line AC ($x + y = 2$) to get $x = 2$, so $C=(2{,}0)$.

Step 2: Calculate the slopes of the sides

Next, we calculate the slopes of the sides AB and AC. The equation of line AB is $3y - x = 2$, which can be rewritten as $y = \frac{1}{3}x + \frac{2}{3}$. So, the slope of AB, $m_{AB}$, is $\frac{1}{3}$. The equation of line AC is $x + y = 2$, which can be rewritten as $y = -x + 2$. So, the slope of AC, $m_{AC}$, is $-1$.

Step 3: Find the equations of altitudes from B and C

The orthocentre P is the intersection of the altitudes. The altitude from B to AC is perpendicular to AC. Since $m_{AC} = -1$, the slope of the altitude from B is $m_{B} = -\frac{1}{-1} = 1$. Using point-slope form with $B(-2{,}0)$, the equation of this altitude is $y - 0 = 1(x - (-2))$, which simplifies to $y = x + 2$. The altitude from C to AB is perpendicular to AB. Since $m_{AB} = \frac{1}{3}$, the slope of the altitude from C is $m_{C} = -\frac{1}{1/3} = -3$. Using point-slope form with $C(2{,}0)$, the equation of this altitude is $y - 0 = -3(x - 2)$, which simplifies to $y = -3x + 6$.

Step 4: Find the coordinates of the orthocentre P

The orthocentre P is the intersection of the two altitudes. We solve the system of equations for the altitudes: $y = x + 2$ and $y = -3x + 6$. Substituting the first into the second gives $x + 2 = -3x + 6$. This simplifies to $4x = 4$, so $x = 1$. Substituting $x=1$ into $y = x + 2$ gives $y = 1 + 2 = 3$. Thus, the orthocentre is $P(1{,}3)$.

Step 5: Calculate the area of triangle PBC

Finally, we calculate the area of triangle PBC. The base BC lies on the $x$ -axis. The length of the base BC is the distance between $B(-2{,}0)$ and $C(2{,}0)$, which is $|2 - (-2)| = 4$ units. The height of the triangle PBC with respect to base BC is the absolute value of the $y$ -coordinate of P, which is $|3| = 3$ units. Using the formula for the area of a triangle, Area $= \frac{1}{2} \times \text{base} \times \text{height}$, we get Area $= \frac{1}{2} \times 4 \times 3 = 6$ square units.

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