We have, $${\text{max}}\left\{ {\left| {x - 1} \right|,\,\left| {y - 2} \right|} \right\} = 4$$
If $$\left\{ {\left| {x - 1} \right| \geqslant \left| {y - 2} \right|} \right\},$$
then $$\left| {x - 1} \right| = 4,$$
i.e., if $$\left( {x + y - 3} \right)\left( {x - y + 1} \right) \geqslant 0,$$
then $$x = - 3{\text{ or }}5$$
If $$\left| {y - 2} \right| \geqslant \left| {x - 1} \right|,$$
then $$\left| {y - 2} \right| = 4$$
i.e., $$\left( {x + y - 3} \right)\left( {x - y + 1} \right) \leqslant 0,$$
then $$y = - 2{\text{ or }}6$$
So, the locus of $$P$$ bounds a square, the equation of whose sides are $$x = - 3,\,x = 5,\,y = - 2,\,y = 6$$
Thus, the area is $${\left( 8 \right)^2} = 64.$$

The image of (0, 0) on $$y-x=0$$   by the line $$2x-y=1$$   is $$\left( {\frac{4}{5},\, - \frac{2}{5}} \right).$$
The point of intersection of $$y-x=0$$   and $$2x-y=1$$   is $$\left( {1,\,1} \right).$$
$$\therefore $$  the required line is the line joining the points $$\left( {1,\,1} \right)$$  and $$\left( {\frac{4}{5},\, - \frac{2}{5}} \right).$$

Since point at 4 units from $$P\left( {2,\,3} \right)$$   will be $$A\left( {4\,\cos \,\theta + 2,\,4\,\sin \,\theta + 3} \right)$$       and this point will satisfy the equation of line $$x+y=7$$
$$ \Rightarrow \cos \,\theta + \sin \,\theta = \frac{1}{2}$$
On squaring
$$\eqalign{ & \Rightarrow \sin \,2\theta - \frac{3}{4} \Rightarrow \frac{{2\,\tan \,\theta }}{{1 + {{\tan }^2}\theta }} = - \frac{3}{4} \cr & \Rightarrow 3\,{\tan ^2}\theta + 8\,\tan \,\theta + 3 = 0 \cr & \Rightarrow \tan \,\theta = \frac{{ - 8 \pm 2\sqrt 7 }}{6}\,\,\,\,\,\,\left( {{\text{ignoring }} - {\text{ve sign}}} \right) \cr & \Rightarrow \tan \,\theta = \frac{{ - 8 + 2\sqrt 7 }}{6} \cr & \Rightarrow \tan \,\theta = \frac{{1 - \sqrt 7 }}{{1 + \sqrt 7 }} \cr} $$

Equation of $$PQ$$  is
$$\frac{x}{h} + \frac{y}{k} = 1.....(1)$$

Since, (1) passes through the fixed point (2, 3) Then,
$$\frac{2}{h} + \frac{3}{k} = 1$$
Then, the locus of $$R$$ is $$\frac{2}{x} + \frac{3}{y} = 1$$   or $$3x+2y=xy.$$

Let the two perpendicular lines be the co-ordinate axes.
Let $$\left( {x,\,y} \right)$$  be the point sum of whose distances from two axes is 1 then we must have
$$\left| x \right| + \left| y \right| = 1\,\,\,\,{\text{or}}\,\, \pm x \pm y = 1$$
These are the four lines $$x+y=1, \,x-y=1, \,-x+y=1, \,-x-y=1$$
Any two adjacent sides are perpendicular to each other. Also each line is equidistant from origin. Therefore figure formed is a square.

The distance between $$\left( {\frac{2}{m},\,2} \right)$$  and $$\left( {\frac{6}{m},\,6} \right)$$  is less than 5
$$\eqalign{ & \Rightarrow {\left( {\frac{2}{m} - \frac{6}{m}} \right)^2} + {\left( {2 - 6} \right)^2} < 25 \cr & \Rightarrow \frac{{16}}{{{m^2}}} < 9 \cr & \Rightarrow {m^2} > \frac{{16}}{9} \cr & \Rightarrow m > \frac{4}{3}{\text{ or }}m < \frac{{ - 4}}{3} \cr} $$

$$S$$ is the midpoint of $$Q$$ and $$R$$
Therefore, $$S \equiv \left( {\frac{{7 + 6}}{2},\,\frac{{3 - 1}}{2}} \right) = \left( {\frac{{13}}{2},\,1} \right)$$

Now slope of $$PS = m = \frac{{2 - 1}}{{2 - \frac{{13}}{2}}} = - \frac{2}{9}$$
Now equation of the line passing through $$\left( {1,\, - 1} \right)$$  and parallel to $$PS$$  is
$$y + 1 = - \frac{2}{9}\left( {x - 1} \right)\,\,\,{\text{or}}\,\,\,2x + 9y + 7 = 0$$

Clearly, the circumcenter of triangle $$ABQ$$  will lie on the perpendicular bisector of line $$AB.$$  Now, the equation of perpendicular bisector of line $$AB$$  is $$3x - 4y + \frac{7}{2} = 0.$$    Hence, the locus of circumcenter is $$6x - 8y + 7 = 0.$$

We have
$$\eqalign{ & 2p = \left| {\frac{{0 + 0 - 1}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}} \right| \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{4{p^2}}} \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{2}{{8{p^2}}} \cr & \Rightarrow \frac{1}{{{a^2}}},\,\frac{1}{{8{p^2}}},\,\frac{1}{{{b^2}}}\,{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow {a^2},\,8{p^2},\,{b^2}{\text{ }}\,{\text{are in H}}{\text{.P}}{\text{.}} \cr} $$

From the figure, $$ - 1 < a < 1.$$