Find the vertex $A$ as intersection of lines $AB$ and $AC$.

Given: $3y-x=2$ $x+y=2$

From the second equation, $x=2-y$

Substitute into the first: $3y-(2-y)=2$ $3y-2+y=2$ $4y=4$ $y=1$ Then, $x=2-1=1$

So, $A=(1,1)$

Find points $B$ and $C$ on the $x$-axis.

Since $B$ lies on line $AB$ and on the $x$-axis, put $y=0$ in $3y-x=2$: $-x=2 \Rightarrow x=-2$ Hence, $B=(-2,0)$

Since $C$ lies on line $AC$ and on the $x$-axis, put $y=0$ in $x+y=2$: $x=2$ Hence, $C=(2,0)$

Equation of side $BC$.

Both $B$ and $C$ lie on the $x$-axis, so $BC: y=0$

Find the altitude from $A$ to $BC$.

Since $BC$ is horizontal, the altitude from $A(1,1)$ is vertical: $x=1$

Find the altitude from $B$ to $AC$.

Line $AC$ is $x+y=2 \Rightarrow y=-x+2$ so its slope is $-1$.

Therefore, a perpendicular line has slope $1$. Passing through $B(-2,0)$: $y-0=1(x+2)$ $y=x+2$

Find orthocentre $P$ as intersection of altitudes.

Intersect $x=1$ with $y=x+2$

gives $y=1+2=3$

So, $P=(1,3)$

Find area of triangle $PBC$.

Points: $B=(-2,0), \quad C=(2,0), \quad P=(1,3)$

Base $BC$ lies on the $x$-axis with length: $BC=2-(-2)=4$

Height from $P$ to $BC$ is the vertical distance from $(1,3)$ to $y=0$, i.e. $3$

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

Compare with stored answer.

Derived answer is $6$, which matches option D.