One side of the triangle is parallel to the $$y$$-axis and another side is parallel to the $$x$$-axis. So, the triangle is a right-angled triangle. Hence, the middle point of the hypotenuse is the circumcenter. Solving $$x=2,\,x+2y=4$$    we get one end of the hypotenuse, and solving $$y+1=0,\,x+2y=4$$     we get the other end. Their coordinates are $$\left( {2,\,1} \right)$$  and $$\left( {6,\, - 1} \right).$$
$$\therefore {\text{circumcentre}} = \left( {\frac{{2 + 6}}{2},\,\frac{{1 - 1}}{2}} \right)$$

Let the two perpendicular lines be the coordinate axes and let the point be $$P\left( {h,\,k} \right).$$
Then sum of the distances of $$P\left( {h,\,k} \right)$$  from the coordinate axes is $$\left| h \right| + \left| k \right|.$$   It is given that $$\left| h \right| + \left| k \right| = 1.$$
So, locus of $$\left( {h,\,k} \right)$$  is $$\left| x \right| + \left| y \right| = 1.$$   This gives four lines $$x + y = 1,\, x - y = 1,\, -x + y = 1,\, -x - y = 1$$         which enclose a square.

From the figure, $$y$$-coordinates of $$P$$ vary from $$0$$ to $$1$$. So, to be an interior point,
$$0<$$  ordinate of $$P<1$$

We have
$$x - y = 4$$   & $$2x + 3y + 7 = 0$$
On solving, we get,
$$x = 1$$  & $$y = - 3$$
(these are coordinates of centre of the circle)
$$ \Rightarrow {\text{ radius}} = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {4 + 3} \right)}^2}} = 5\sqrt 2 $$

$$3x+4y = 0$$    is one of the lines of the pair
$$\eqalign{ & 6{x^2} - xy + 4c{y^2} = 0, \cr & {\text{Put}}\,\,y = - \frac{3}{4}x, \cr & {\text{we get}}\,\,6{x^2} + \frac{3}{4}{x^2} + 4c{\left( { - \frac{3}{4}x} \right)^2} = 0 \cr & \Rightarrow 6 + \frac{3}{4} + \frac{{9c}}{4} = 0\,\, \Rightarrow c = - 3\, \cr} $$

$$\eqalign{ & \left( {1 + OB} \right)\cos \,{30^ \circ } = 2\sqrt 3 \cr & \Rightarrow 1 + OB = 2\sqrt 3 \times \frac{2}{{\sqrt 3 }} = 4 \cr & \therefore OB = 3 \cr & \therefore B = \left( {3,\,0} \right){\text{ and }}'m'{\text{ of }}BC = \tan \,{150^ \circ } = - \frac{1}{{\sqrt 3 }} \cr & \therefore {\text{the equation of }}BC{\text{ is }}y - 0 = - \frac{1}{{\sqrt 3 }}\left( {x - 3} \right) \cr} $$

$$\because $$ $$A$$ is the mid point of $$PQ,$$  therefore
$$\frac{{a + 0}}{2} = 3,\,\frac{{0 + b}}{2} = 4\,\, \Rightarrow a = 6,\,\,b = 8$$
$$\therefore $$ Equation of line is $$\frac{x}{6} + \frac{y}{8} = 1{\text{ or }}4x + 3y = 24$$

We have the equation
$$\eqalign{ & 2{x^2} - xy - {y^2} = 0 \cr & \Rightarrow \left( {2x + y} \right)\left( {x - y} \right) = 0 \cr} $$
If $$\left( {h,\,k} \right)$$  be the point then remaining pair is $$\left( {2x + y + h} \right)\left( {x - y + k} \right) = 0$$
Where, $$2x + y + h = 0$$    and $$x - y + k = 0$$
It passes through the point $$\left( {1,\,0} \right)$$
$$\eqalign{ & \therefore \,2 \times 1 + 0 + h = 0 \Rightarrow 2 + h = 0 \Rightarrow h = - 2 \cr & {\text{and }}1 - 0 + k = 0 \Rightarrow 1 + k = 0 \Rightarrow k = - 1 \cr} $$
$$\therefore $$  Required pair is
$$\eqalign{ & \left( {2x + y - 2} \right)\left( {x - y - 1} \right) = 0 \cr & \Rightarrow 2{x^2} - 2xy - 2x + xy - {y^2} - y - 2x + 2y + 2 = 0 \cr & \therefore \,2{x^2} - xy - {y^2} - 4x + y + 2 = 0 \cr} $$

$$\eqalign{ & \left( {x - 1} \right)\left( {x - 3} \right) + \left( {y - 2} \right)\left( {y - 4} \right) = 0 \cr & \Rightarrow AC \bot BC\,\,\,\,\,\, \Rightarrow \angle ACB = {90^ \circ } \cr & \therefore C{\text{ is on the circle whose diameter is }}AB \cr & AB = 2\sqrt 2 \cr & {\text{As ar }}\left( {\Delta ABC} \right) = 1,\,\frac{1}{2}.2\sqrt 2 .\left( {{\text{altitude}}} \right) = 1 \cr & \therefore {\text{altitude}} = \frac{1}{{\sqrt 2 }}{\text{ < radius}}{\text{.}} \cr & {\text{So, there are four possible positions of }}C. \cr & {\bf{Note:}} \cr & {\text{If ar}}\left( {\Delta ABC} \right) = 2,{\text{ altitude}} = {\text{radius }}\, \Rightarrow {\text{ two possible positions of }}C. \cr} $$

\[\begin{array}{l} {x_2} = {x_1}r,\,{x_3} = {x_1}{r^2}{\rm{ \,and \,so\, }}{y_2} = {y_1}r,\,{y_3} = {y_1}{r^2}\\ \Delta = \left| \begin{array}{l} {x_1}\,\,\,{y_1}\,\,\,1\\ {x_2}\,\,\,{y_2}\,\,\,1\\ {x_3}\,\,\,{y_3}\,\,\,1 \end{array} \right| = r.\,{r^2}\left| \begin{array}{l} {x_1}\,\,\,{y_1}\,\,\,1\\ {x_1}\,\,\,{y_1}\,\,\,1\\ {x_1}\,\,\,{y_1}\,\,\,1 \end{array} \right| = 0 \end{array}\]
Hence the points lie on a line, i.e., they are collinear.