The distance between the two lines represented by the equation $9{x^2} - 24xy + 16{y^2} - 12x + 16y - 12 = 0$
a. $\dfrac{8}{5}$
b. $\dfrac{6}{5}$
c. $\dfrac{{11}}{5}$
d. None of these

Hint: We will simplify the given equation by taking terms common. Then, substitute $3x - 4y$ by $t$ and factorise the equation. Put each factor equal to 0 and form two equations of line. Determine the distance between the two parallel lines by using the formula, if $ax + by + {c_1} = 0$ and $ax + by + {c_2} = 0$ are two parallel lines, then the distance between them is given by $\dfrac{{\left| {{c_2} - {c_1}} \right|}}{{\sqrt {{a^2} + {b^2}} }}$

Complete step by step answer:

We will first separate the equation of two lines from the given equation of both the lines.
We are given that both the equations of line is represented by $9{x^2} - 24xy + 16{y^2} - 12x + 16y - 12 = 0$
Rewrite the first three terms of the equation to form an expression of the whole square.
Here, we can see that
${\left( {3x} \right)^2} - 2\left( {3x} \right)\left( {4y} \right) + {\left( {4y} \right)^2} - 12x + 16y - 12 = 0$
Now, we know that ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$
Therefore, we have
${\left( {3x - 4y} \right)^2} - 12x + 16y - 12 = 0$
Now we can take 4 common from second and third term of the above equation.
${\left( {3x - 4y} \right)^2} - 4\left( {3x - 4y} \right) - 12 = 0$
Let $3x - 4y$ be represented by $t$
Then, we have
${t^2} - 4t - 12 = 0$
We will factorise the above equation by splitting the middle term
$
  {t^2} - 6t + 2t - 12 = 0 \\
   \Rightarrow t\left( {t - 6} \right) + 2\left( {t - 6} \right) = 0 \\
   \Rightarrow \left( {t + 2} \right)\left( {t - 6} \right) = 0 \\
$
Put back the value of $t$
Then,
$\left( {3x - 4y + 2} \right)\left( {3x - 4y - 6} \right) = 0$
Put each factor equals to 0 to get two equations of the line.
Hence, we have equations of line as
$3x - 4y + 2 = 0$ and $3x - 4y - 6 = 0$
Since, the coefficient of $x$ and $y$ are same for both the equations, the equations are parallel to each other.
Therefore, we have to find the distance between two parallel lines.
As it is known that is $ax + by + {c_1} = 0$ and $ax + by + {c_2} = 0$ are two parallel lines, then the distance between them is given by $\dfrac{{\left| {{c_2} - {c_1}} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
Thus, the distance between $3x - 4y + 2 = 0$ and $3x - 4y - 6 = 0$ is calculated as,
$\dfrac{{\left| { - 6 - 2} \right|}}{{\sqrt {{3^2} + {4^2}} }} = \dfrac{8}{{\sqrt {9 + 16} }} = \dfrac{8}{{\sqrt {25} }} = \dfrac{8}{5}$
Hence, option a is correct.

Note: Parallel lines have the same slope, thus the ratio of coefficients of $x$ and $y$ of two parallel lines is the same. Also, parallel lines never intersect each other, and the distance between any two sets of corresponding points of parallel lines is always the same.