Point of intersection of $$3x - 4y - 7 = 0$$ and

$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the

circle and radius $$=7$$

$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$

$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$

As per question area of one sector $$=3$$ area of another sector

$$ \Rightarrow $$ at center by one sector $$ = 3 \times $$ angle at center by another sector

Let one angle be $$\theta $$ then other $$=30$$

Clearly $$\theta + 3\theta = 180 \Rightarrow \theta = {45^ \circ }$$

$$\therefore$$ Angle between the diameters represented by combined equation

$$a{x^2} + 2\left( {a + b\,\,\,xy} \right) + b{y^2} = 0$$ is $${45^ \circ }$$

$$\therefore$$ Using $$tan$$ $$\theta $$ $$ = {{2\sqrt {{h^2} - ab} } \over {a + b}}$$

we get $$\tan \,{45^ \circ } = {{2\sqrt {{{\left( {a + b} \right)}^2} - ab} } \over {a + b}}$$

$$ \Rightarrow 1 = {{2\sqrt {{a^2} + {b^2} + ab} } \over {a + b}}$$

$$ \Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right)$$

$$ \Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab$$

$$ \Rightarrow 3{a^2} + 3{b^2} + 2ab = 0$$

$${s_1} = {x^2} + {y^2} + 2ax + cy + a = 0$$

$${s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0$$

Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is

given by $${s_1} - {s_2} = 0$$

$$ \Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$

Given that $$5x + by - a = 0$$ passes through $$P$$ and $$Q$$

$$\therefore$$ The two equations should represent the same line

$$ \Rightarrow {a \over 1} = {{c - d} \over b} = {{a + 1} \over { - a}}$$

$$ \Rightarrow a + 1 = - {a^2}$$

$${a^2} + a + 1 = 0$$

No real value of $$a.$$

Let the center be $$\left( {\alpha ,\beta } \right)$$

As It cuts the circle $${x^2} + {y^2} = {p^2}$$ orthogonally

$$\therefore$$ Using $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},\,\,$$ we get

$$2\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0$$

$$ = {c_1} - {p^2} \Rightarrow {c_1} = {p^2}$$

Let equation of circle is

$${x^2} + {y^2} - 2\alpha x - 2\beta y + {p^2} = 0$$

It passes through

$$\left( {a,b} \right) \Rightarrow {a^2} + {b^2} - 2\alpha a - 2\beta b + {p^2} = 0$$

$$\therefore$$ Locus of $$\left( {\alpha ,\beta } \right)$$ is

$$\therefore$$ $$2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0.$$