The given lines are $$y + 2x = 1$$   and $$y + 2x = 2.$$
The distance between the lines is $$\frac{{\left( {2 - 1} \right)}}{{\sqrt 5 }} = \frac{1}{{\sqrt 5 }}.$$
The side length of the triangle is $$\frac{1}{{\sqrt 5 }}{\text{cosec}}\,{60^ \circ } = \frac{2}{{\sqrt 5 }}$$

The given line passes through $$\left( { - 3,\,5} \right)$$  and $$\left( {2,\,0} \right).$$  Its equation is

$$\eqalign{ & y - {y_1} = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right) \cr & \Rightarrow \left( {y - 5} \right) = \left( {\frac{{0 - 5}}{{2 + 3}}} \right)\left( {x + 3} \right) \cr & \Rightarrow y = - x + 2......\left( 1 \right) \cr} $$
Slope $$ = m = -1$$   and slope of perpendicular line $$ = - \frac{1}{m} = 1$$
Equation of this line passing through $$\left( {3,\,3} \right)$$  is :
$$\eqalign{ & \left( {y - 3} \right) = 1\left( {n - 3} \right) \cr & \Rightarrow y = x \cr} $$
From equation $$\left( 1 \right)$$  we get,
$$\eqalign{ & x = - x + 2 \cr & \Rightarrow x = 1{\text{ and }}y = 1 \cr} $$

Here $$x + y + \lambda \left( {2x - y + 1} \right) = 0$$
$$ \Rightarrow $$ each line passes through the point of intersection of the lines $$x+y=0$$   and $$2x-y+1=0,$$    which is $$\left( { - \frac{1}{3},\,\frac{1}{3}} \right)$$
The required line passes through $$\left( { - \frac{1}{3},\,\frac{1}{3}} \right)$$   and is perpendicular to the line joining $$\left( {1,\,4} \right)$$  and $$\left( { - \frac{1}{3},\,\frac{1}{3}} \right).$$
$$\therefore \,'m'$$  of the required line $$ = \frac{{ - 1}}{{\frac{{4 - \frac{1}{3}}}{{1 + \frac{1}{3}}}}} = - \frac{4}{{11}}$$
$$\therefore $$ the required line is $$y - \frac{1}{3} = - \frac{4}{{11}}\left( {x + \frac{1}{3}} \right){\text{ or }}12x + 33y = 7$$

$$\eqalign{ & 3a + {a^2} - 2 = 0 \cr & \Rightarrow {a^2} + 3a - 2 = 0; \cr & \Rightarrow a = \frac{{ - 3 \pm \sqrt {9 + 8} }}{2} = \frac{{ - 3 \pm \sqrt {17} }}{2} \cr} $$

$$\eqalign{ & \tan \left( {{{180}^ \circ } - \theta } \right) = {\text{slope of }}AB = - 3 \cr & \therefore \tan \,\theta = 3 \cr & \therefore \frac{{OC}}{{AC}} = \tan \,\theta ,\,\frac{{OC}}{{BC}} = \cot \,\theta \cr & \Rightarrow \frac{{BC}}{{AC}} = \frac{{\tan \,\theta }}{{\cot \,\theta }} = {\tan ^2}\theta = 9 \cr} $$

Let $$P, \,O, \,R,$$   be the vertices of $$\Delta PQR$$

Since $$PS$$  is the median, $$S$$ is mid-point of $$QR$$
$${\text{So, }}S = \left( {\frac{{7 + 6}}{2},\,\frac{{3 - 1}}{2}} \right) = \left( {\frac{{13}}{2},\,1} \right)$$
Now, slope of $$PS\,\,\,\,\, = \frac{{2 - 1}}{{2 - \frac{{13}}{2}}} = - \frac{2}{9}$$

Since, required line is parallel to $$PS$$  therefore slope of required line $$=$$ slope of $$PS$$
Now, equation of line passing through $$\left( {1,\, - 1} \right)$$  and having slope $$ - \frac{2}{9}$$ is
$$\eqalign{ & y - \left( { - 1} \right) = - \frac{2}{9}\left( {x - 1} \right) \cr & \Rightarrow 9y + 9 = - 2x + 2 \cr & \Rightarrow 2x + 9y + 7 = 0 \cr} $$

$$\eqalign{ & {x^2} - {y^2} + 2y = 1 \cr & \Rightarrow x = \pm \left( {y - 1} \right) \cr} $$

Bisectors of above lines are $$x = 0$$  and $$y= 1$$

So area between $$x = 0, \,y= 1$$    and $$x + y =3$$   is shaded region shown in figure.
Area $$ = \frac{1}{2} \times 2 \times 2 = 2\,{\text{sq}}{\text{.}}\,{\text{units}}$$

$$\eqalign{ & x = \frac{{a\,\cos \,t + b\,\sin \,t + 1}}{3} \Rightarrow a\,\cos \,t + b\,\sin \,t = 3x - 1 \cr & y = \frac{{a\,\sin \,t - b\,\cos \,t}}{3} \Rightarrow a\,\sin \,t - b\,\cos \,t = 3y \cr & {\text{Squaring & adding, }}{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2} \cr} $$

$${\left( {{x^2} - {a^2}} \right)^2}{\left( {{x^2} - {b^2}} \right)^2} + {c^4}{\left( {{y^2} - {a^2}} \right)^2} = 0$$
This being the sum of two perfect squares, each term must be zero.
Hence, we get
$$\eqalign{ & {\left( {{x^2} - {a^2}} \right)^2}{\left( {{x^2} - {b^2}} \right)^2} = 0 \cr & {\text{or }}\left( {{x^2} - {a^2}} \right)\left( {{x^2} - {b^2}} \right) = 0 \cr & {\text{or }}\left( {x - a} \right)\left( {x + a} \right)\left( {x - b} \right)\left( {x + b} \right) = 0......\left( 1 \right) \cr & {\text{and }}{c^4}{\left( {{y^2} - {a^2}} \right)^2} = 0 \cr & {\text{or }}{c^2}\left( {{y^2} - {a^2}} \right) = 0 \cr & {\text{or }}{c^2}\left( {y + a} \right)\left( {y - a} \right) = 0......\left( 2 \right) \cr} $$
Equation number $$\left( 1 \right)$$ holds good for $$x = \pm a{\text{ or }}x = \pm b$$
Equation number $$\left( 2 \right)$$ is satisfied by $$y = \pm a$$
As both of these should be simultaneously satisfied, the given equation represents $$8$$ points which we get as a result of different combinations of $$\left( 1 \right)$$ and $$\left( 2 \right),$$ namely $$\left( { \pm a,\, \pm a} \right),\,\left( { \pm b,\, \pm a} \right).$$

The lines are $$\left( {\sqrt 3 x - y} \right)\left( {x - \sqrt 3 y} \right) = 0$$
Their separate equations are $$y = \tan \,{60^ \circ }.x,\,y = \tan \,{30^ \circ }.x$$
After rotation, the separate equations are $$y = \tan \,{90^ \circ }.x,\,y = \tan \,{60^ \circ }.x$$
$$\therefore $$ the combined equation in the new position is $$x\left( {\sqrt 3 x - y} \right) = 0$$