Ler P(h, k) be a variable point moving in such a way that the square of its distance from A(3, -2) is numerically equal to its distance from the line 5x - 12y = 13.
\(\therefore \) (h-3)² + (k+2)² = \(\frac { \left| 5h-12k-13 \right| }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right) }^{ 2 } } } \)
\(\Rightarrow \) 13{(h - 3)² + (k + 2)²} =  \(\pm \) (5h - 2k - 13)
\(\Rightarrow \) 13(h² + k²) - 83h + 64k + 182 = 0
or  13(h² + k²) - 73h + 40k + 156 = 0
Hence, the locus of (h, k) is
13 (x² +y²) - 83x + 64 y +182 =0
or 13 (x² + y²) - 73 + 40y + 156 =0

Let slope of the line be m and the coordinates of fixed point P be (x₁, y₁).
Then, equation of line is y - y₁ = m (x - x₁)    ....(i)
Let the given points be A(2, 0), B(0, 2) and C(1,1).
Now, perpendicular distance from A
\(=\frac { 0-{ y }_{ 1 }-m\left( 2-{ x }_{ 1 } \right) }{ \sqrt { 1+{ m }^{ 2 } } } \)
Perpendicular distance from B \(=\frac { 2-{ y }_{ 1 }-m\left( 0-{ x }_{ 1 } \right) }{ \sqrt { 1+{ m }^{ 2 } } } \)
and perpendicular distance from C \(=\frac { 1-{ y }_{ 1 }-m\left( 1-{ x }_{ 1 } \right) }{ \sqrt { 1+{ m }^{ 2 } } } \)
Now,\(\frac { \left[ -{ y }_{ 1 }-2m+{ mx }_{ 1 }+2-{ y }_{ 1 }+{ mx }_{ 1 }+1-{ y }_{ 1 }-m+{ mx }_{ 1 } \right] }{ \sqrt { 1+{ m }^{ 2 } } } =0\)
\(\Rightarrow \) -3y₁ - 3m +3mx₁ + 3 = 0
\(\Rightarrow \) -y₁ - m + mx₁ + 1 = 0
\(\Rightarrow \)  y₁ =  -m + mx₁ + 1
On substituting this value in Eq. (i), we gwt
y + m - mx₁ - 1 = mx - mx₁ \(\Rightarrow \) y - 1 = m(x - 1)
Thus,(x₁, y₁) = (1, 1)

Given equation of line is
\(\frac { x }{ a } +\frac { y }{ b } =1\) ....(i)
Perpendicular length from the origin to the line (i) is
\(p=\frac { 1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } =\frac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \Rightarrow { p }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } \)
Since, a² , p² and b² are in AP.
\(\therefore { 2p }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\Rightarrow \frac { 2{ a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } ={ a }^{ 2 }{ +b }^{ 2 }\)
\(\Rightarrow \) 2a²b² = (a² + b²)² \(\Rightarrow \)2a²b² = a⁴+b⁴+2a²b²
\(\Rightarrow \) a⁴ + b⁴ = 0

Given point is \((a \ { cos }^{ 3 }\theta ,a sin^{ 3 }\ \theta )\) and the line is \(x \sec \theta+y \operatorname{cosec} \theta=a\)
\(\because\)  Slope of given line \(=-\frac { sec\theta }{ cosec\theta } =-tan\theta \)
\(\therefore\)  Slope of required line = \(\frac { 1 }{ tan\theta } =cot\theta \)
Now, equation of the required line having slope cot \(\theta\) and passing through \((a \ { cos }^{ 3 }\theta ,a sin^{ 3 }\ \theta )\), is
\(y-a \sin ^{3} \theta=\cot \theta\left(x-a \cos ^{3} \theta\right) \)
\(\Rightarrow y-a \sin ^{3} \theta=\frac{\cos \theta}{\sin \theta}\left(x-a \cos ^{3} \theta\right) \)
\(\Rightarrow y \sin \theta-a \sin ^{4} \theta=x \cos \theta-a \cos ^{4} \theta \)
\(\Rightarrow x \cos \theta-y \sin \theta=a \cos ^{4} \theta-a \sin ^{4} \theta \)
\(\Rightarrow x \cos \theta-y \sin \theta \)
\(=a\left[\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\right] \)
\(\Rightarrow x \cos \theta-y \sin \theta=a \cos 2 \theta\)

Let the coordinates of the point A be ( x, 0 ) .From the figure, the slope of the reflected ray is given by
tan \(\theta \) = \(\frac{3}{5 - x} \) .... ( i )
 

Again , the slope of the incident ray is given by 
tan ( \(\pi\) - \(\theta\) ) = \(\frac{-2}{ x-1 }\)   [ \(\because\) m = \(\frac{y_2-y_1}{x_2-x_1}\) ]
\(\Rightarrow\) - tan \(\theta\)  = \( \frac{-2}{ x-1 }\) [ \(\because\) tan ( \(\pi\) - \(\theta\) = - tan \(\theta\)  )  ]
\(\Rightarrow\)  tan \(\theta\) =  \( \frac{2}{ x - 1}\)
From Eqs.(i) and (ii), we get
\(\frac{3}{ 5 - x }\) = \(\frac{2}{ x - 1}\)
\(\Rightarrow\) 3x  - 3 = 10 -2x
\(\Rightarrow\)  x = \( \frac{13}{5}\)
Therefore , the required coordinates of the point A are ( \(\frac{13}{5}\),0 ).