Step 1: Use the right-angle condition.
The triangle is right-angled at $B(4,k)$, so $AB\perp BC$, meaning $m_{AB}\cdot m_{BC}=-1$.
Step 2: Write the two slopes.
$m_{AB}=\frac{k-1}{4-2}=\frac{k-1}{2}$ and $m_{BC}=\frac{4-k}{3-4}=\frac{4-k}{-1}$.
Step 3: Solve for k.
$\frac{k-1}{2}\cdot\frac{4-k}{-1}=-1$ gives $(k-1)(4-k)=2$, so $-k^2+5k-4=2$, i.e. $k^2-5k+6=0$, $(k-2)(k-3)=0$. Since $k$ is not odd, $k=2$, so $B=(4,2)$.
Step 4: Locate the orthocentre.
In a right triangle the orthocentre is the right-angle vertex, so $H=(4,2)$.
Step 5: Locate the circumcentre.
The circumcentre is the midpoint of the hypotenuse $AC$: $O=\left(\frac{2+3}{2},\frac{1+4}{2}\right)=\left(\frac52,\frac52\right)$.
Step 6: Equation through H and O.
Slope $=\frac{2-\frac52}{4-\frac52}=\frac{-1/2}{3/2}=-\frac13$. The line $y-2=-\frac13(x-4)$ simplifies to $x+3y=10$.
\[ \boxed{x+3y=10} \]