Given equation is
3x² + 7xy + 2y² + 5x + 5y + 2 = 0
Factorizing 3x² + 7xy + 2y² = (x + 2y) (3x + y)
\(\therefore \) 3x² + 7xy + 2y² + 5x + 5y + 2 = (x + 2y + l)(3x + y + m)
Equating the x and y Co-ordinates both sides,

We get 5 = m + 3l  ...(1)
5 = 2m + l   ....(2)

Substituting l = 1 in (2) we get,
5 = 2m + 1 \(\Rightarrow \) 2m = 4 \(\Rightarrow \) m = 2.
Hence the separate equations are
x + 2y + 1 = 0 and 3x + y + 2 = 0.

let p(x₁,y₁) be any point on the locus ​\(\therefore\)​  \(\sqrt { { \left( { x }_{ 1 }-4 \right) }^{ 2 }+{ \left( { y }_{ 1 }-0 \right) }^{ 2 } } =\frac { 1 }{ 2 } \left| \frac { { x }_{ 1 }-16 }{ \sqrt { { 1 }^{ 2 }+{ 0 }^{ 2 } } } \right| \)
\(\Rightarrow\)\(\sqrt { { \left( { x }_{ 1 }-4 \right) }^{ 2 }+{ y }_{ 1 }^{ 2 } } =\frac { 1 }{ 2 } \left| \frac { { x }_{ 1 }-16 }{ \sqrt { { 1 }^{ 2 }+{ 0 }^{ 2 } } } \right| \)
Squaring  both sides, (x₁ - 4)² + y₁² = \(\frac{1}{4}\)(x₁ - 16)²
⇒ 4[x₁² + 16 - 8 x₁ + y₁²] = x₁² - 32x₁ + 256
⇒ 3x₁² + 4y₁² - 192 = 0
Locus of (x₁, y₁) is 3x² + 4y² = 192

Solution: Let p (x₁,y₁) be any point on the locus, such that the square of its distance from A (3, -2) is equal to its distance from 5x - 12y = 13.
\(\therefore\) (x₁ - 3)² + (y₁ + 2)² = \(\frac { \left| { 5x }_{ 1 }-12{ y }_{ 1 }+13 \right| }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right) }^{ 2 } } } \)
⇒ 13[(x₁ - 3)² + (y₁ + 2)²] = 土(5x₁ - 12y₁ + 13)
⇒ 13(x₁² - 6x₁ + 9 + y₁² + 4 + 4y₁) = 士(5x₁ - 12y₁ + 13)
Case (i)
⇒ 13(x₁² + y₁² - 6x₁ + 4y₁ + 13) = 5x₁ - 12y₁ + 13
⇒ 13x₁² + 13y₁² - 83x₁ + 64y₁ + 182 = 0
Case (ii)
13(x₁² + y₁² - 6x₁ + 4y₁ + 13) = -(5x₁ - 12y₁ + 13)
13x₁² + 13y₁² - 73x₁ + 40y₁ + 156 = 0
\(\therefore\) Locus of (x₁, y₁) is 13x² + 13y² - 83x + 64y + 182 = 0 (or) 13x² + 13y² - 73x + 40y + 156 = 0

Let p(x,y) be any point on the parabola whose focus is F(-3, 2) and the directrix is x + y - 4 = 0.
Draw pm perpendicular to x + y - 4 = 0
Then FP = pm \(\Rightarrow\) FP² = pm²
\(\Rightarrow { (x+3) }^{ 2 }+{( y-2) }^{ 2 }={ \left[ \frac { x+y-4 }{ \sqrt { 1+1 } } \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+6x+9+{ y }^{ 2 }-4y+4=\frac { { x }^{ 2 }+{ y }^{ 2 }+16+2xy-8x-8y }{ 2 } \)
\(\Rightarrow\) 2(x² + y² + 6x - 4y + 13) = x² + y² + 2y - 8x - 8y + 16
\(\Rightarrow\) x² + y² - 2xy + 20x + 10 = 0.

The centre is the point of intersection of the diameters.
Solving 2x - 3y + 12 = 0 ....(1)  and
x + 4y - 5 = 0  ....(2)
(1)     \(\rightarrow\)    2x - 3y + 12 = 0
                           -      -      +
(2)\(\times\)2 \(\rightarrow\)   2x + 8y - 10 = 0
                   ___________________
                       -11y + 22 = 0
\(\Rightarrow \) -11y = -22 
\(\Rightarrow \) y = 2
Substituting y = 2 in (2) we get,
x + 4(2) - 5 = 0 
\(\Rightarrow \) x + 8 - 5 = 0 
\(\Rightarrow \) x + 3 = 0 
\(\Rightarrow \) x = -3.
\(\therefore \)  (-3, 2) is the center of the circle.
Also, given area = 154  \(\Rightarrow \) \(\pi\)r² = 154
\(\Rightarrow \frac { 22 }{ 7 } \times { r }^{ 2 }=5\)
\(\Rightarrow { r }^{ 2 }=\frac { 154\times 7 }{ 22 } =\frac { 14\times 7 }{ 2 } \)
\(\Rightarrow\) r² = 49  \(\Rightarrow\) r = 7.
\(\therefore \)  Equation of the circle is (x + 3)² + (y - 2)² = 49
\(\Rightarrow\)  x² + 6x + 9 y² - 4y + 4 = 49
\(\Rightarrow\)  x² + y² + 6x - 4y - 36 = 0.