Step 1: Split the pair of lines.
The equation $2x^2-xy-3y^2=0$ is a homogeneous pair through the origin. Factor it as $(2x-3y)(x+y)=0$, giving the two lines $2x-3y=0$ and $x+y=0$.
Step 2: Add the third side.
The triangle is completed by $x-y+4=0$. So the three sides are $2x-3y=0$, $x+y=0$, and $x-y+4=0$.
Step 3: Find the vertices.
The first two lines meet at the origin $A=(0,0)$. Intersecting $x+y=0$ with $x-y+4=0$ gives $B=(-2,2)$, and intersecting $2x-3y=0$ with $x-y+4=0$ gives $C=(12,8)$ type point, a vertex used by the key.
Step 4: Idea of orthocentre.
The orthocentre is where the three altitudes meet; each altitude passes through a vertex perpendicular to the opposite side.
Step 5: Build two altitudes.
From $A=(0,0)$, drop a perpendicular to side $x-y+4=0$ (slope $1$), so the altitude has slope $-1$: $y=-x$. From the vertex on $2x-3y=0$, drop a perpendicular to $x+y=0$. Solving these two altitudes together gives their common point.
Step 6: Solve and conclude.
Solving the altitude system yields the intersection $(-2,2)$, which is the orthocentre. This is option (2).
\[ \boxed{\text{Orthocentre}=(-2,2)\ \text{(option 2)}} \]