$${L_1}:y - x = 0$$

$${L_2}:2x + y = 0$$

$${L_3}:y + 2 = 0$$

On solving the equation of line $${L_1}$$ and $${L_2}$$ we get their point of

intersection $$(0, 0)$$ i.e., origin $$O.$$

On solving the equation of line $${L_1}$$ and $${L_3},$$

we get $$P=(-2, -2).$$

Similarly, we get $$Q = \left( { - 1, - 2} \right)$$

We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [ Angle Bisector Theorem of a Triangle ]

$$\therefore$$ $${{PR} \over {RQ}} = {{OP} \over {OQ}} = {{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} } \over {\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} }}$$

$$ = {{2\sqrt 2 } \over {\sqrt 5 }}$$

Slope of line $$L = - {b \over 5}$$

Slope of line $$K = - {3 \over c}$$

Line $$L$$ is parallel to line $$k.$$

$$ \Rightarrow {b \over 5} = {3 \over c} \Rightarrow bc = 15$$

$$(13,32)$$ is a point on $$L.$$

$$\therefore$$ $${{13} \over 5} + {{32} \over b} = 1 \Rightarrow {{32} \over b} = - {8 \over 5}$$

$$ \Rightarrow b = - 20 \Rightarrow c = - {3 \over 4}$$

Equation of $$K:$$ $$y - 4x = 3$$

$$\,\,\,\,\,\,\,\,\,\,\,$$ $$ \Rightarrow 4x - y + 3 = 0$$

Distance between $$L$$ and $$K$$

$$ = {{\left| {52 - 32 + 3} \right|} \over {\sqrt {17} }} = {{23} \over {\sqrt {17} }}$$

Given that

$$P\left( {1,0} \right),Q\left( { - 1,0} \right)$$

and $${{AP} \over {AQ}} = {{BP} \over {BQ}} = {{CP} \over {CQ}} = {1 \over 3}$$

$$ \Rightarrow 3AP = AQ$$

$$\,\,\,\,\,\,$$ Let $$A = (x,y)$$ then $$3AP = AQ \Rightarrow 9A{P^2} = A{Q^2}$$

$$ \Rightarrow 9{\left( {x - 1} \right)^2} + 9{y^2} = {\left( {x + 1} \right)^2} + y{}^2$$

$$ \Rightarrow 9{x^2} - 18x + 9 + 9{y^2} = {x^2} + 2x + 1 + {y^2}$$

$$ \Rightarrow 8{x^2} - 20x + 8{y^2} + 8 = 0$$

$$ \Rightarrow {x^2} + {y^2} - {5 \over 3}x + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\therefore$$ A lies on the circle given by eq. $$(1).$$ As $$B$$ and $$C$$

also follow the same condition, - they must lie on the same circle.

$$\therefore$$ Center of circumcircle of $$\Delta ABC$$

$$=$$ Center of circle given by $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$

If the lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$

and $${\left( {{p^2} + 1} \right)^2}x + \left( {{p^2} + 1} \right)y + 2q = 0$$

are perpendicular to a common line then these lines -

must be parallel to each other,

$$\therefore$$ $${m_1} = {m_2} \Rightarrow - {{p\left( {{p^2} + 1} \right)} \over { - 1}} = - {{{{\left( {{p^2} + 1} \right)}^2}} \over {{p^2} + 1}}$$

$$ \Rightarrow \left( {{p^2} + 1} \right)\left( {p + 1} \right) = 0$$

$$ \Rightarrow p = - 1$$

$$\therefore$$ $$p$$ can have exactly one value.