Using slopes to find orthocentre of a triangle

Question

Let vertices of a triangle be $A(1,0)$, $B(-2,1)$, $C(5,2)$. Let orthocentre be $H(a,b)$. Slope of $AH=b/(a-1)$, slope of $BH = (b-1)/(a+2)$, slope of $CH = (b-2)/(a-5)$. I assumed $AH, BH, CH$ to be mutually perpendicular. Slopes of perpendicular lines have relation $p*q=-1$ where p and q are slopes of the lines. Using the above relation I got $a^2+b^2-a-b-2=0$, $a^2+b^2-6a-2b+5=0$ and $a^2+b^2-3a-3b-8=0$. On solving the three equations i got $a=2$ and $b=-7$ which are the correct coordinates. Even though I took $AH$, $BH$, $CH$ to be mutually perpendicular, which is completely false, I still got the right answer. How?