Let (1, 0) be the vertex and x + y – 7 = 0 be the equation of the latus rectum of a parabola and if the equation of parabola is $x^2+y^2-2xy+px+qy+r=0,$ then (p + q + r) is equal to

Focus is foot of perpendicular drawn from vertex on latus rectum i.e. (4, 3) and equation of the directrix is x + y + c = 0 i.e $\left|\frac{1+c}{\sqrt{2}}\right|=3\sqrt{2}\Rightarrow c=5or-7$

$\therefore$ equation of the directrix is x + y + 5 = 0

$\therefore$ equation of the required parabola is ${(x-4)}^2+{(y-3)}^2={\left(\frac{x+y+5}{\sqrt{2}}\right)}^2\Rightarrow x^2+y^2-2xy-26x-22y+25=0$

Compare with $x^2+y^2-2xy+px+qy+r=0\Rightarrow p=-26,q=-22,r=25$

$\therefore p+q+r=-23$