If a triangle $ABC$ has vertices $A(1,-6), B(2,-3)$ and $C(3,-2)$, then the coordinates of its orthocenter are....

Let the vertices of the triangle be $A(1, -6)$, $B(2, -3)$, and $C(3, -2)$.

The slope of the side $BC$ is $m_{BC} = \dfrac{-2 - (-3)}{3 - 2} = 1$.
The altitude from $A$ to $BC$ is perpendicular to $BC$, so its slope is $-1$.
The equation of the altitude from $A(1, -6)$ is:
$y - (-6) = -1(x - 1)$
$x + y = -5$ ... (i)

The slope of the side $AC$ is $m_{AC} = \dfrac{-2 - (-6)}{3 - 1} = \dfrac{4}{2} = 2$.
The altitude from $B$ to $AC$ is perpendicular to $AC$, so its slope is $-\dfrac{1}{2}$.
The equation of the altitude from $B(2, -3)$ is:
$y - (-3) = -\dfrac{1}{2}(x - 2)$
$2y + 6 = -x + 2$
$x + 2y = -4$ ... (ii)

The orthocenter is the point of intersection of the altitudes. Solving equations (i) and (ii):
Subtracting (i) from (ii), we get:
$(x + 2y) - (x + y) = -4 - (-5)$
$y = 1$

Substituting $y = 1$ into equation (i):
$x + 1 = -5$
$x = -6$

The coordinates of the orthocenter are $(-6, 1)$.

Answer: $(-6, 1)$