$\text{ Consider the family of lines }(x-y-6)+\lambda (2x+y+3)=0\text{ and }(x+2y-4)+\mu (3x-2y-4)=0$.If the lines of these two families are at right angle to each other, then the locus of their point of intersection, is 

$solving x-y-6=0 and 2x+y+3=0 we get (1,-5)\text{ and }x+2y-4=0 , 3x-2y-4=0 we get (2,1)$

 Given family of lines are concurrent at A(1, -5) and B(2, 1). Now point P(h, k) is such that AP $\perp$ BP 

$\begin{array}{l}\Rightarrow \left(\frac{k+5}{h-1}\right)\cdot \left(\frac{k-1}{h-2}\right)=-1\\ \therefore \text{ Locus of }(h,k)\text{ is }{x}^2+y^2-3x+4y-3=0.\end{array}$