Step 1: Recognise the structure of the equation.
The given equation is $4x^2 + 20xy + 25y^2 + 2x + 5y - 12 = 0$. Notice that $4x^2 + 20xy + 25y^2 = (2x + 5y)^2$. The equation has a perfect square in the quadratic part.
Step 2: Rewrite using $u = 2x + 5y$.
Let $u = 2x + 5y$. Also $2x + 5y$ appears in the linear terms too: $2x + 5y = u$. So the equation becomes $u^2 + u - 12 = 0$.
Step 3: Solve the quadratic in $u$.
$u^2 + u - 12 = 0$. Factorising: $(u + 4)(u - 3) = 0$, giving $u = -4$ or $u = 3$.
Step 4: Convert back to equations of lines.
$u = 2x + 5y$, so: $2x + 5y = -4$, i.e. $2x + 5y + 4 = 0$, and $2x + 5y = 3$, i.e. $2x + 5y - 3 = 0$. These are two parallel lines (same coefficients of $x$ and $y$).
Step 5: Apply the distance formula for parallel lines.
For two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$, the distance is $\frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$. Here $a = 2$, $b = 5$, $c_1 = 4$, $c_2 = -3$.
Step 6: Compute the distance.
\[ \text{Distance} = \frac{|4 - (-3)|}{\sqrt{2^2 + 5^2}} = \frac{7}{\sqrt{4 + 25}} = \frac{7}{\sqrt{29}} \]
Step 7: State the final answer.
\[ \boxed{\dfrac{7}{\sqrt{29}}} \]