The combined equation is $$\left( {x + y - 1} \right)\left( {x - y - 1} \right) = 0.$$     After rotation, the lines are $$x = 1$$  and $$y = \sqrt 3 \left( {x - 1} \right).$$   So, the combined equation after rotation is $$\left( {x - 1} \right)\left( {\sqrt 3 x - y - \sqrt 3 } \right) = 0.$$

In an equilateral triangle the orthocenter and the centroid are the same.
$$OPQ$$  is the equilateral triangle in which $$OC \bot PQ.$$
Clearly, the point $$H$$ which divides $$OC$$  internally in the ratio 2 : 1 is the orthocenter.
$$\eqalign{ & {\text{Clearly, }}OC = \frac{1}{{\sqrt 2 }}{\text{.}}\,\,{\text{So,}}\,\,OH = \frac{2}{3} \times \frac{1}{{\sqrt 2 }} \cr & \therefore H = \left( {\frac{2}{{3\sqrt 2 }}\cos \,{{45}^ \circ },\frac{2}{{3\sqrt 2 }}\sin \,{{45}^ \circ }} \right) \cr} $$

Let any line through the origin meets the given lines at $$E$$ and $$F$$ as shown in figure

Now, from the figure, triangles $$OAE$$   and $$OCF$$   are similar.
Therefore,
$$\frac{{OE}}{{OF}} = \frac{{OA}}{{OC}} = \frac{{\frac{9}{4}}}{3} = \frac{3}{4}$$

The point $$R$$ should be such that $$PR$$  is reflected along $$RQ$$  from the line $$y=0.$$
The equation of $$P'Q$$  ( where $$P'$$ is the image of $$P$$ ) is $$3x-2y=5.$$
$$R$$ is the point of intersection of $$y=0$$  and $$3x-2y=5.$$

As $$L$$ has intercepts $$a$$ and $$b$$ on axes, equation of $$L$$ is
$$\frac{x}{a} + \frac{y}{b} = 1.....(1)$$
Let $$x$$ and $$y$$ axes be rotated through an angle $$\theta $$ in anticlockwise direction.
In new system intercepts are $$p$$ and $$q,$$ therefore equation of $$L$$ becomes
$$\frac{x}{p} + \frac{y}{q} = 1.....(2)$$
KEY CONCEPT : As the origin is fixed in rotation, the distance of line from origin in both the cases should be same.
$$\eqalign{ & \therefore {\text{We get }}d = \left| {\frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}} \right| = \left| {\frac{1}{{\sqrt {\frac{1}{{{p^2}}} + \frac{1}{{{q^2}}}} }}} \right| \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{{p^2}}} + \frac{1}{{{q^2}}} \cr & \therefore \,\,({\bf{B}})\,{\text{is the correct answer}}{\text{.}} \cr} $$

$$\because $$  Coordinates of $$Q$$ are $$\left( {3,\,0} \right)$$  & it passes through $$PQ.$$
$$\therefore $$  Putting the values of $$\left( {x = 3} \right)\& \left( {y = 0} \right)$$    in options we get :
Equation of line $$PQ = 7x + y - 21 = 0$$

Co-ordinates of $$A = \left( {a\,\cos \,\alpha ,\,a\,\sin \,\alpha } \right)$$

Equation of OB,
$$\eqalign{ & y = \tan \left( {\frac{\pi }{4} + \alpha } \right)x \cr & {\text{CA}}{ \bot ^{\text{r}}}{\text{OB}} \cr & \therefore \,{\text{slope of CA }} = - \cot \left( {\frac{\pi }{4} + \alpha } \right) \cr} $$
Equation of CA
$$\eqalign{ & y - a\,\sin \,\alpha = - \cot \left( {\frac{\pi }{4} + \alpha } \right)\left( {x - a\,\cos \,\alpha } \right) \cr & \Rightarrow \left( {y - a\,\sin \,\alpha } \right)\left( {\tan \left( {\frac{\pi }{4} + \alpha } \right)} \right) = \left( {a\,\cos \,\alpha - x} \right) \cr & \Rightarrow \left( {y - a\,\sin \,\alpha } \right)\left( {\frac{{\tan \frac{\pi }{4} + \tan \,\alpha }}{{1 - \tan \frac{\pi }{4}\tan \,\alpha }}} \right) = \left( {a\,\cos \,\alpha - x} \right) \cr & \Rightarrow \left( {y - a\,\sin \,\alpha } \right)\left( {1 + \tan \,\alpha } \right) = \left( {a\,\cos \,\alpha - x} \right)\left( {1 - \tan \,\alpha } \right) \cr & \Rightarrow \left( {y - a\,\sin \,\alpha } \right)\left( {\cos \,\alpha + \,\sin \,\alpha } \right) = \left( {a\,\cos \,\alpha - x} \right)\left( {\cos \,\alpha - \sin \,\alpha } \right) \cr & \Rightarrow y\left( {\cos \,\alpha + \sin \,\alpha } \right) - a\,\sin \,\alpha \,\cos \,\alpha - a\,{\sin ^2}\alpha = a\,{\cos ^2}\alpha - a\,\cos \,\alpha \,\sin \,\alpha - x\,\left( {\cos \,\alpha - a\,\sin \,\alpha } \right) \cr & \Rightarrow y\left( {\cos \,\alpha + \,\sin \,\alpha } \right) + x\left( {\cos \,\alpha - \,\sin \,\alpha } \right) = a \cr & y\left( {\,\sin \,\alpha + \,\cos \,\alpha } \right) + x\left( {\cos \,\alpha - \,\sin \,\alpha } \right) = a \cr} $$

$$\eqalign{ & C = \left( {2 + 1.\cos \,{{60}^ \circ },\,1.\sin \,{{60}^ \circ }} \right) = \left( {\frac{5}{2},\,\frac{{\sqrt 3 }}{2}} \right) \cr & E = \left( {1,\,1.\sin \,{{60}^ \circ } + 1.\sin \,{{60}^ \circ }} \right) = \left( {1,\,\sqrt 3 } \right) \cr & \therefore {\text{the equation of }}CE{\text{ is }}y - \sqrt 3 = \frac{{\sqrt 3 - \frac{{\sqrt 3 }}{2}}}{{1 - \frac{5}{2}}}.\left( {x - 1} \right) \cr} $$

For the concurrency of three lines,
$$\eqalign{ & a\left[ {\left( {m + 1} \right)b - \left( {m + 2} \right)c} \right] - ma\left( {b - c} \right) + \left( {m + 2} \right)bc - \left( {m + 1} \right)bc = 0 \cr & \Rightarrow \frac{1}{c} - \frac{1}{b} - \frac{1}{b} + \frac{1}{a} = \frac{1}{c} + \frac{1}{a} - \frac{2}{b} = 0 \cr & \therefore \,\frac{1}{a},\,\frac{1}{b},\,\frac{1}{c}{\text{ are in A}}{\text{.P}}{\text{., for all }}m \cr & \therefore \,a,\,b,\,c\,{\text{are in H}}{\text{.P}}{\text{., for all }}m. \cr} $$

$$\because \,AB = BC = CA = 4\sqrt 5 ,$$
i.e., given triangle is equilateral.
(Incentre of a triangle are same as the centroid when triangle is equilateral)
Hence, incentre
$$\eqalign{ & = \left( {\frac{{7 - 1 + 3 + 2\sqrt 3 }}{3},\,\frac{{1 + 5 + 3 + 4\sqrt 3 }}{3}} \right) \cr & = \left( {3 + \frac{2}{{\sqrt 3 }},\,3 + \frac{4}{{\sqrt 3 }}} \right) \cr} $$