Show that the following equation represents a pair of lines and find distance between them.
I. $ 8{x^2} - 24xy + 18{y^2} - 6x + 9y - 5 = 0 $
II. $ 9{x^2} - 6xy + {y^2} + 18x - 6y + 8 = 0 $
III. $ {x^2} - 2\sqrt 3 xy + 3{y^2} - 3x + 3\sqrt 3 y - 4 = 0 $

Hint: The general equation representing a pair of straight lines is given as $ a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 $ . To find if the given pairs represent a pair of parallel lines or not, we will convert the given equations to the general form if they are not already in the general form. Then we will compare the given equations with the general form and find the value of all the coefficients. The condition for an equation to represent a pair of parallel lines is –
 $ {h^2} = ab $ and $ a{f^2} = b{g^2} $ . On putting the values of h, a, b, f and g in the given conditions, we can find the correct answers.

Complete step-by-step answer:
I.$ 8{x^2} - 24xy + 18{y^2} - 6x + 9y - 5 = 0 $
We see that all the equations are already in the general form.
On comparing with the general form, we get –
 $ a = 8,\,h = - 12,\,b = 18,\,g = - 3,\,f = - \dfrac{9}{2},\,c = - 5 $
Condition for the given equation to represent a pair of parallel lines is –
 $ {h^2} = ab $ and $ a{f^2} = b{g^2} $
 $
  {( - 12)^2} = 8 \times 18\,\,and\,\,8 \times {(\dfrac{{ - 9}}{2})^2} = 18 \times {( - 3)^2} \\
   \Rightarrow 144 = 144\,\,and\,\,162 = 162 \\
  $
As the given equation satisfies the condition, so it represents a pair of parallel lines.
Distance between the pair of parallel lines is given by the formula –
 $ 2\sqrt {\dfrac{{{g^2} - ac}}{{a(a + b)}}} $
 $
   \Rightarrow distance = 2\sqrt {\dfrac{{{{( - 3)}^2} - 8( - 5)}}{{8(8 + 18)}}} \\
   \Rightarrow distance = 2\sqrt {\dfrac{{9 + 40}}{{208}}} = 2\sqrt {\dfrac{{49}}{{208}}} = \dfrac{{2 \times 7}}{{4\sqrt {13} }} \\
   \Rightarrow distance = \dfrac{7}{{2\sqrt {13} }} \\
  $

II.$ 9{x^2} - 6xy + {y^2} + 18x - 6y + 8 = 0 $
On comparing with the general form, we get –
 $ a = 9,\,h = - 3,\,b = 1,\,g = 9,\,f = - 3,\,c = 8 $
Condition for the given equation to represent a pair of parallel lines is –
 $ {h^2} = ab $ and $ a{f^2} = b{g^2} $
 $
  {( - 3)^2} = 9 \times 1\,\,and\,\,9 \times {( - 3)^2} = 1 \times {(9)^2} \\
   \Rightarrow 9 = 9\,\,and\,\,81 = 81 \\
  $
As the given equation satisfies the condition, so it represents a pair of parallel lines.
Distance between the pair of parallel lines is given by the formula – $ 2\sqrt {\dfrac{{{g^2} - ac}}{{a(a + b)}}} $
 $
   \Rightarrow distance = 2\sqrt {\dfrac{{{{(9)}^2} - 9(8)}}{{9(9 + 1)}}} \\
   \Rightarrow distance = 2\sqrt {\dfrac{{81 - 72}}{{90}}} = 2\sqrt {\dfrac{9}{{90}}} \\
   \Rightarrow distance = \dfrac{2}{{\sqrt {10} }} \\
  $

III. $ {x^2} - 2\sqrt 3 xy + 3{y^2} - 3x + 3\sqrt 3 y - 4 = 0 $
On comparing with the general form, we get –
 $ a = 1,\,h = - \sqrt 3 ,\,b = 3,\,g = - \dfrac{3}{2},\,f = - \dfrac{{3\sqrt 3 }}{2},\,c = - 4 $
Condition for the given equation to represent a pair of parallel lines is –
 $ {h^2} = ab $ and $ a{f^2} = b{g^2} $
 $
  {( - \sqrt 3 )^2} = 1 \times 3\,\,and\,\,1 \times {(\dfrac{{ - 3\sqrt 3 }}{2})^2} = 3 \times {( - \dfrac{3}{2})^2} \\
   \Rightarrow 3 = 3\,\,and\,\,\dfrac{{27}}{4} = \dfrac{{27}}{4} \\
  $
As the given equation satisfies the condition, so it represents a pair of parallel lines.
Distance between the pair of parallel lines is given by the formula – $ 2\sqrt {\dfrac{{{g^2} - ac}}{{a(a + b)}}} $
 $
   \Rightarrow distance = 2\sqrt {\dfrac{{{{( - \dfrac{3}{2})}^2} - 1( - 4)}}{{1(1 + 3)}}} \\
   \Rightarrow distance = 2\sqrt {\dfrac{{\dfrac{9}{4} + 4}}{4}} = 2\sqrt {\dfrac{{25}}{4}} = 2 \times \dfrac{5}{2} \\
   \Rightarrow distance = 5 \;
  $

Note: Lines are said to be parallel when they are extended up to infinity and don’t intersect at any point. The slope of all the parallel lines is equal. When two lines are parallel to each other, they form a pair of parallel lines. Thus, all the equations that satisfy the given condition represent a pair of parallel lines.