Being a pair of lines, $$abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0.$$       This gives $$m=4.$$

If the lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$      and $${\left( {{p^2} + 1} \right)^2}x + \left( {{p^2} + 1} \right)y + 2q = 0$$       are perpendicular to a common line then these lines must be parallel to each other,
$$\eqalign{ & \therefore {m_1} = {m_2} \Rightarrow - \frac{{p\left( {{p^2} + 1} \right)}}{{ - 1}} = - \frac{{{{\left( {{p^2} + 1} \right)}^2}}}{{{p^2} + 1}} \cr & \Rightarrow \left( {{p^2} + 1} \right)\left( {p + 1} \right) = 0 \cr & \Rightarrow p = - 1 \cr} $$
$$\therefore p$$  can have exactly one value.

Let the vertex $$C$$ be $$\left( {h,\,k} \right),$$   then the centroid of $$\Delta ABC$$   is $$\left( {\frac{{2 - 2 + h}}{3},\,\frac{{ - 3 + 1 + k}}{3}} \right)\,\,{\text{or }}\,\left( {\frac{h}{3},\,\frac{{ - 2 + k}}{3}} \right)$$
It lies on $$2x+3y=1$$
$$\eqalign{ & \Rightarrow \frac{{2h}}{3} - 2 + k = 1 \cr & \Rightarrow 2h + 3k = 9 \cr} $$
$$=$$ Locus of $$C$$ is $$2x+3y=9$$

Let the joining points be $$A\left( {1,\,1} \right)$$   and $$B\left( {2,\,4} \right)$$
Let point $$C$$ divides line $$AB$$  in the ratio 3 : 2
So, by section formula we have
$$C = \left( {\frac{{3 \times 2 + 2 \times 1}}{{3 + 2}},\,\frac{{3 \times 4 + 2 \times 1}}{{3 + 2}}} \right) = \left( {\frac{8}{5},\,\frac{{14}}{5}} \right)$$
Since Line $$2x + y = k$$   passes through $$C\left( {\frac{8}{5},\,\frac{{14}}{5}} \right)$$
$$\therefore \,\,C$$  satisfies the equation $$2x + y=k$$
$$ \Rightarrow \frac{{2 + 8}}{5} + \frac{{14}}{5} = k\,\,\,\,\, \Rightarrow k = 6$$

As the points are in order, the area
\[\begin{array}{l} = \left| {\frac{1}{2}\left\{ {\left| \begin{array}{l} 4\,\,\,1\\ 3\,\,\,6 \end{array} \right| + \left| \begin{array}{l} \,\,\,\,3\,\,\,\,6\\ - 5\,\,\,1 \end{array} \right| + \left| \begin{array}{l} - 5\,\,\,\,\,\,\,\,1\\ - 3\,\, - 3 \end{array} \right| + \left| \begin{array}{l} - 3\,\,\, - 3\\ - 3\,\,\,\,\,\,\,\,\,0 \end{array} \right| + \left| \begin{array}{l} - 3\,\,\,\,0\\ \,\,\,\,4\,\,\,\,\,1 \end{array} \right|} \right\}} \right|\\ = 30\,{\rm{uni}}{{\rm{t}}^2} \end{array}\]

The length of the perpendicular from the origin to the line in both the cases is the same. So
$$\eqalign{ & \frac{{\frac{0}{a} + \frac{0}{b} - 1}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }} = \frac{{\frac{0}{p} + \frac{0}{q} - 1}}{{\sqrt {\frac{1}{{{p^2}}} + \frac{1}{{{q^2}}}} }} \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{{p^2}}} + \frac{1}{{{q^2}}}. \cr} $$

If variable point is $$P$$ and $$S\left( { - 2,\,0} \right)$$   then $$PS = \frac{2}{3}PM$$   where $$PM$$ is the perpendicular distance of point $$P$$ from given line $$x = - \frac{9}{2}$$
$$\therefore $$ By definition $$P$$ describes an ellipse. $$\left( {e = \frac{2}{3} < 1} \right)$$

$$\eqalign{ & {\text{The area of the triangle}} = \frac{1}{2} \times \left( {x{\text{ - intercept}}} \right) \times \left( {y{\text{ - intercept}}} \right) \cr & = \frac{1}{2} \times \left( { - \frac{c}{a}} \right) \times \left( { - \frac{c}{b}} \right) \cr & = \frac{1}{2}.\frac{{{c^2}}}{{ab}} \cr & = \frac{1}{2}\,\,\,\,\left( {\because \,\,a,\,c,\,b{\text{ are in GP}}} \right). \cr} $$

Area of $$\Delta OAB = S = \frac{1}{2}ab......\left( {\text{i}} \right)$$
Equation of $$AB$$  is $$\frac{x}{a} + \frac{y}{b} = 1$$
Putting $$\left( {\alpha ,\,\beta } \right),$$  we get
$$\eqalign{ & \frac{\alpha }{a} + \frac{\beta }{b} = 1 \cr & \Rightarrow \frac{\alpha }{a} + \frac{{a\beta }}{{2S}} = 1\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{using }}\left( {\text{i}} \right)} \right] \cr & \Rightarrow {a^2}\beta - 2aS + 2\alpha S = 0 \cr & \therefore \,a\, \in \,R \Rightarrow D \geqslant 0 \cr & 4{S^2} - 8\alpha \beta S \geqslant 0 \cr & \Rightarrow S \geqslant 2\alpha \beta \cr} $$
Least value of $$S = 2\alpha \beta $$

$$3x + 4y + 5 = 0{\text{ or }}y = \frac{{ - 3}}{4}x + \frac{{ - 5}}{4}$$
Slope $$ = \frac{{ - 3}}{4}$$
Slope of required line, $$m = \frac{{ - 1}}{{ - \frac{3}{4}}} = \frac{4}{3}$$
Also line passes through $$\left( {4,\, - 5} \right)$$
Equation of line
$$\eqalign{ & y + 5 = \frac{4}{3}\left( {x - 4} \right) \cr & \Rightarrow 3y + 15 = 4x - 16 \cr & \Rightarrow 4x - 3y - 31 = 0 \cr} $$