3x²+10xy+8y²+14x+22y+15=0
a= 3, h = 5, b = 8, g = 7, f= 11, c = 15
The condition is af²+bg²+ch² abc-2fgh = 0
3(11)² +8(7)² +15(5)² -(3)(8)(15)-2(11)(7)(5)=363+392+375-360-770=0
Hence the equation represents a pair of straight lines
tanθ=\(\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\pm \frac { 2 }{ 11 } \)
tanθ=\(\frac { 2 }{ 11 } \)
⇒ θ=tan-1\(\left( \frac { 2 }{ 11 } \right) \).

9x² +24xy+16y² +21x+28y+6=0
\(\left| \begin{matrix} 2h=24 \\ h=12 \end{matrix} \right| b=16\left| \begin{matrix} 2g=21 \\ g=\frac { 21 }{ 2 } \end{matrix} \right| c=6\left| \begin{matrix} 2f=28 \\ f=14 \end{matrix} \right| \)
h²-ab=(12)² -9(16)=144-144=0
∴ The lines are parallel.
9x²+24xy+16y² =(3x+4y)(3x+4y)
Let 9x² + 24xy+16y²+21x+28y+6 = (3x+4y+l)(3x+4y+m)
Equating the coefficients of x and constant term
3l+ 3m =21
lm= 6
Solving we get, l=1 or 6 m = 6
m = 6 or 1
∴ The separate equations are 3x + 4y + 1 = 0 and 3x + 4y + 6 =0
The distance between the parallel lines are \(\left| \frac { 6-1 }{ \sqrt { 9+16 } } \right| =\frac { 5 }{ 5 } \)=1 unit.

Equation of line through the intersection of straight lines 5x - 6y = 1 and 3x + 2y + 5
5x - 6y - 1 + k (3x + 2y + 5) = 0
x (5 + 3k) + y (-6 + 2k) + (-1 + 5k) = 0
This is perpendicular to 3x - 5y + 11 = 0
That is, the product of their slopes is -1
-\(\left( \frac { 5+3k }{ -6+2k } \right) \left( -\frac { 3 }{ -5 } \right) \)=-1
⇒ \(\frac { 15+9k }{ -30+10k } \)=1
45=k
Required equation is 5x - 6y -1 + 45 (3x + 2y + 5) = 0
140x + 84y + 224 = 0
20x + 12y + 32 = 0
5x+ 3y+ 8 = 0.

Given that:  OM = 4 units.
\(\angle BAX=120^o\)
\(\therefore \angle BAO=180^o-120^o\) or \(\angle MAO =60^o\)
\(\angle MOA+\angle MAO=90^o\)     [\(\therefore OM\bot AB\)]
\(\theta + 60^o=90^o\quad \quad \therefore \theta=30^o\)
So, equation of AB in its normal form
\(x\cos\theta+y\sin\theta=p\)
\(\Rightarrow x\cos30^o+y\sin30^o=4\)
\(\Rightarrow x\times \frac{\sqrt{3}}{2}+y\times \frac{1}{2}=4\Rightarrow \sqrt{3}x+y=8\)

Slope of the line joining the points (2, 3) and (3, -1) is
\(\frac { -1-3 }{ 3-2 } \)=-4
Slope of the required line which is perpendicular to it
=\(\frac { -1 }{ -4 } =\frac { 1 }{ 4 } \)   [∵ m₁m₂=-1]
Equation of the line passing through the point (5, 2) is
y-2 =\(\frac { 1 }{ 4 } \)(x-5)    [y-y₁=m(x-x₁)]
⇒ 4y-8=x-5
⇒ x-4y+3=0