Let coordinate of point R = (h, k).

Equation of line PQ,

(y $$-$$ 3) = m (x $$-$$ 2).

Put y = 0 to get coordinate of point p,

0 $$-$$ 3 = (x $$-$$ 2)

$$ \Rightarrow $$ x = 2 $$-$$ $${3 \over m}$$

$$\therefore\,\,\,$$ p = (2 $$-$$ $${3 \over m}$$, 0)

As p = (h, 0) then

h = 2 $$-$$ $${3 \over m}$$

$$ \Rightarrow $$ $${3 \over m}$$ = 2 $$-$$ h

$$ \Rightarrow $$ m = $${3 \over {2 - h}}$$ . . . . . . (1)

Put x = 0 to get coordinate of point Q,

y $$-$$ 3 $$=$$ m (0 $$-$$ 2)

$$ \Rightarrow $$ y = 3 $$-$$ 2m

$$\therefore\,\,\,$$ point Q = (0, 3 $$-$$ 2m)

And From the graph you can see Q = (0, k).

$$\therefore\,\,\,$$ k = 3 $$-$$ 2m

$$ \Rightarrow $$ m = $${{3 - k} \over 2}$$ . . . . (2)

By comparing (1) and (2) get

$${3 \over {2 - h}} = {{3 - k} \over 2}$$

$$ \Rightarrow $$ (2 $$-$$ h)(3 $$-$$ k) = 6

$$ \Rightarrow $$ 6 $$-$$ 3h $$-$$ 2K + hk = 6

$$ \Rightarrow $$ 3 h + 2K = hk

$$\therefore\,\,\,$$ locus of point R is 3x + 2y= xy