If two vertices of a triangle are $(3,-2)$ and $(-2,3)$ and its orthocentre is $(-6,1)$. Then the difference between ordinate and abscissa of the third vertex of the triangle is

Let the vertices of the triangle be $A(3, -2)$, $B(-2, 3)$, and $C(x, y)$. The orthocentre $H$ is given as $(-6, 1)$.

The slope of $BC$ is $m_{BC} = \dfrac{y - 3}{x - (-2)} = \dfrac{y - 3}{x + 2}$. Since $AH \perp BC$, the slope of $AH$ is $m_{AH} = \dfrac{1 - (-2)}{-6 - 3} = \dfrac{3}{-9} = -\dfrac{1}{3}$.

The condition $m_{AH} \times m_{BC} = -1$ implies $\left(-\dfrac{1}{3}\right) \times \left(\dfrac{y - 3}{x + 2}\right) = -1$, which simplifies to $\dfrac{y - 3}{x + 2} = 3$, so $y - 3 = 3x + 6$, or $3x - y + 9 = 0$.

The slope of $AC$ is $m_{AC} = \dfrac{y - (-2)}{x - 3} = \dfrac{y + 2}{x - 3}$. Since $BH \perp AC$, the slope of $BH$ is $m_{BH} = \dfrac{1 - 3}{-6 - (-2)} = \dfrac{-2}{-4} = \dfrac{1}{2}$.

The condition $m_{BH} \times m_{AC} = -1$ implies $\left(\dfrac{1}{2}\right) \times \left(\dfrac{y + 2}{x - 3}\right) = -1$, which simplifies to $\dfrac{y + 2}{x - 3} = -2$, so $y + 2 = -2x + 6$, or $2x + y - 4 = 0$.

Solving the system of equations $3x - y = -9$ and $2x + y = 4$:

Adding the two equations gives $5x = -5$, so $x = -1$.

Substituting $x = -1$ into $2x + y = 4$ gives $2(-1) + y = 4$, so $y = 6$.

The third vertex $C$ is $(-1, 6)$. The difference between the ordinate and the abscissa is $y - x = 6 - (-1) = 7$.

Answer: $7$