Clearly, $$A$$ will remain as $$\left( {0,\,0} \right),{f_1}$$   will make $$B$$ as $$\left( {0,\,4} \right),{f_2}$$   will make it $$\left( {12,\,4} \right),$$  and $${f_3}$$ will make it $$\left( {4,\,8} \right),{f_1}$$   will make $$C$$ as $$\left( {2,\,4} \right),{f_2}$$   will make it $$\left( {14,\,4} \right),$$  and $${f_3}$$ will make it $$\left( {5,\,9} \right).$$
Finally, $${f_1}$$ will make $$D$$ as $$\left( {2,\,0} \right),{f_2}$$   will make it $$\left( {2,\,0} \right),$$  and $${f_3}$$ will make it $$\left( {1,\,1} \right).$$
So, we finally get $$A\left( {0,\,0} \right),\,B\left( {4,\,8} \right),\,C\left( {5,\,9} \right){\text{ and }}D\left( {1,\,1} \right).$$
Hence,
$$\eqalign{ & {m_{AB}} = \frac{8}{4},\,{m_{BC}} = \frac{{9 - 8}}{{5 - 4}} = 1,\,{m_{CD}} = \frac{{9 - 1}}{{5 - 1}} = \frac{8}{4}, \cr & {m_{AD}} = 1,\,{m_{AC}} = \frac{9}{5},\,{m_{BD}} = \frac{{8 - 1}}{{4 - 1}} = \frac{7}{3} \cr} $$
Hence, the final figure will be a parallelogram.

Clearly from the figure, the reflected ray moves along the $$x$$-axis.

$$\eqalign{ & AB = \sqrt {{{\left( {4 + \,1} \right)}^2} + \,{{\left( {0 + 1} \right)}^2}} = \sqrt {26} \cr & BC = \sqrt {{{\left( {3 + \,1} \right)}^2} + \,{{\left( {5 + 1} \right)}^2}} = \sqrt {52} \cr & CA = \sqrt {{{\left( {4 - 3} \right)}^2} + \,{{\left( {0 - 5} \right)}^2}} = \sqrt {26} \cr} $$
In isosceles triangle side $$AB = CA$$
For right angled triangle, $$B{C^2} = A{B^2} + A{C^2}$$
So, here
$$\eqalign{ & BC = \sqrt {52} \, \cr & {\text{or}}\,{\text{ }}B{C^2} = 52 \cr & {\text{or}}\,{\text{ }}{\left( {\sqrt {26} } \right)^2} + {\left( {\sqrt {26} } \right)^2} = 52 \cr} $$
So, the given triangle is right angled and also isosceles .

Let the equation of the line $$L$$ be $$y - 2 = m\left( {x - 8} \right),\,m < 0$$
Coordinates of $$P$$ and $$Q$$ are $$P\left( {8 - \frac{2}{m},\,0} \right)$$   and $$Q\left( {0,\,2 - 8m} \right)$$
$$\eqalign{ & {\text{So, }}OP + OQ = 8 - \frac{2}{m} + 2 - 8m = 10 + \frac{2}{{ - m}} + 8\left( { - m} \right) \cr & \geqslant 10 + 2\sqrt {\frac{2}{{ - m}} + 8\left( { - m} \right)} \geqslant 18 \cr} $$
absolute min. value of $$OP + OQ = 18.$$

$$\eqalign{ & \tan \,\theta = \sqrt 3 \, \Rightarrow \theta = {60^ \circ }\,\, \Rightarrow \angle PQR = {120^ \circ } \cr & \Rightarrow {\text{bisector will have slope tan }}{120^ \circ } \cr & \Rightarrow {\text{equation of bisector is }}\sqrt 3 x + y = 0 \cr} $$

The given points are $$A\left( { - a, - b} \right),\,B\left( {0,\,0} \right),\,C\left( {a,\,b} \right){\text{ and }}D\left( {{a^2},\,ab} \right)$$
$$\eqalign{ & {\text{Slope of }}AB = \frac{b}{a} = {\text{ slope of }}BC = {\text{ slope of }}BD \cr & \therefore A,\,B,\,C,\,D\,\,\,{\text{are collinear}}{\text{.}} \cr} $$

If $$a,\,b$$  are intercepts then the equation of the line is $$\frac{x}{a} + \frac{y}{b} = 1.$$   It passes through $$\left( {2,\,2} \right).$$
$$\eqalign{ & \therefore \,\frac{2}{a} + \frac{2}{b} = 1.\,\,{\text{Also}}\frac{1}{2}ab = \pm \lambda \,\,\,\,{\text{or }}ab = \pm 2\lambda \cr & \therefore \,\frac{{a + b}}{{ab}} = \frac{1}{2}\,\,\,{\text{or}}\,\frac{{a + b}}{{ \pm 2\lambda }} = \frac{1}{2}\,\,\,{\text{or }}a + b = \pm \lambda \cr} $$
$$\therefore $$  the required equation is $${x^2} - \lambda x + 2\lambda = 0{\text{ or }}{x^2} + \lambda x - 2\lambda = 0$$
Combining these, $${x^2} - \left| \lambda \right|x + 2\left| \lambda \right| = 0.$$

$$\eqalign{ & 2{x^2} - xy - {y^2} + x + 2y - 1 = 0 \cr & \Rightarrow \,\left( {2x + y - 1} \right)\left( {x - y + 1} \right) = 0 \cr} $$
So, the sides $$x-y+1=0$$   and $$x+y+1=0$$   are perpendicular to each other.
$$\therefore $$ the orthocenter is the intersection of the perpendicular sides.

The length of perpendicular from $$P\left( {2,\, - 3} \right)$$   on the given family of lines
$$\eqalign{ & = \frac{{a\left( {4 - 3 + 4} \right) + b\left( {2 + 6 - 3} \right)}}{{\sqrt {{{\left( {2a + b} \right)}^2} + {{\left( {a - 2b} \right)}^2}} }} = \pm \sqrt {10} \,\,\,\,\,\,\,\left( {{\text{given}}} \right) \cr & \Rightarrow 5a + 5b = \pm \sqrt {10\left( {5{a^2} + 5{b^2}} \right)} \cr & \Rightarrow 25{\left( {a + b} \right)^2} = 50\left( {{a^2} + {b^2}} \right) \cr & \Rightarrow 25{\left( {a - b} \right)^2} = 0 \cr & \Rightarrow a = b \cr} $$

Area of triangle $$\Delta AOB = \frac{1}{2} \times \frac{c}{b} \times \frac{c}{a} = \frac{{{c^2}}}{{2ab}}$$

Total area
$$\eqalign{ & = 4 \times {\text{area }}\Delta AOB \cr & = 4 \times \frac{{{c^2}}}{{2ab}} \cr & = \frac{{2{c^2}}}{{ab}} \cr} $$