$d_1:ax+2y+4=0$; $d_2:3x-4y+12=0$; $|d_1d_2|=?$ 
$d_1:ax+2y+4=0$
$d_2:3x-4y+12=0$
$d_1$ // $d_2$
$|d_1d_2|=?$

This is a slightly modified version of a university entrance examination problem set up by the experts at this specific university itself.
I don't know what is meant by $|d_1d_2|$. I searched the internet for what could have been meant by this expression, but I couldn't find it. When taken as the distance between two parallel lines, the answer can be found to be $4$.
I want to object to the problem for the ambiguity on what is meant by $|d_1d_2|$. During the rush of the exam, I thought the problem asked for the absolute value of the multiplication of the two lines, for whatever it would be.
Is this expression ,$|d_1d_2|$, algebraically correct and does it mean the distance between two parallel lines?
 A: You can safely skip the computations in 1. and go directely to 2. if you are only interested in the notation. 


*

*Distance between $d_1$ and $d_2$. Two straight lines represented by the equations 
$$Ax+Bx+C=0\qquad\text{ and } \qquad A'x+B'x+C'=0$$
are parallel if and only if 
$$ AB'=A'B .$$
This means that $$d_1:ax+2y+4=0, \qquad \text{ with }\qquad A=a, B=2, C=4$$  and 
$$d_2:3x-4y+12=0\qquad \text{ with }\qquad A'=3, B'=-4, C'=12$$
are parallel if and only if $$ a(-4)=3(2)\iff a=-\frac{3}{2} .$$
Now, let $d$ be a straight line with equation 
$$d:\tilde{A}x+\tilde{B}y+\tilde{C}=0.$$
This line $d$ is perpendicular to e.g. $d_1$ if and only if 
$$A\tilde{A}+B\tilde{B}=0\iff \tilde{A}=-\frac{B\tilde{B}}{A} ,\qquad \text { 
 with   } A=a=-\frac{3}{2}, B=2,$$
which means that 
$$d: \tilde{A}x+\tilde{B}y+\tilde{C}=0\iff -\frac{B\tilde{B}}{A}x+\tilde{B}y+\tilde{C}=0 \iff Bx-Ay-\frac{A\tilde{C}}{\tilde{B}}=0.$$
Hence $d$ is represented by
$$d: 2x+\dfrac{3}{2}y+\frac{3}{4}\tilde{C}=0.$$
For $\tilde{C}=0 $, we get a line $d'$ parallel to $d$:
$$d': 2x+\dfrac{3}{2}y=0.$$
If we compute the coordinates of the intersection points of $d_1$ and $d_2$  with $d'$, we find that 
$$
\left\{ 
\begin{align}
-\frac{3}{2}x+2y+4&=0 \\[2ex] 
2x+\frac{3}{2}y&=0
\end{align}
\right. \iff \left\{ 
\begin{array}{l}
x=\dfrac{24}{25} \\[2ex]
y=-\dfrac{32}{25}
\end{array}
\right. 
$$
and
$$
\left\{ 
\begin{align}
3x-4y+12&=0 \\[2ex] 
2x+\frac{3}{2}y&=0
\end{align}
\right. \iff \left\{ 
\begin{array}{l}
x=-\dfrac{36}{25} \\[2ex] 
y=\dfrac{48}{25}.
\end{array}
\right. 
$$
As you claimed the distance between $d_1$ and $d_2$ is indeed $4$, because
$$\sqrt{\left( \frac{24}{25}+\frac{36}{25}\right) ^{2}+\left( -\frac{32}{25}-
\frac{48}{25}\right) ^{2}}=4.$$


*Notation


*

*As for the notation of the length of the line segment $AB$, that is the distance between two points $A$ and $B$ is normally written $\overline{AB}$. However, $\overline{AB}$  may also denote the line segment itself (here) instead of the length of the line segment, which is then denoted by $AB$. Another notation for the line segment is  $[A,B]$ (here). The symbol $|\ldots|$ may denote Euclidean norm or Euclidean length or magnitude (here)

*Concerning the notation $|d_1d_2|$, I've never seen it before, but from the context, the answer to your last question "does it mean the distance between [the given] parallel lines?" seems reasonable to assume so.



A: Since $d_1||d_2$ we have $-{a\over 2} = {3\over 4}$ so $a= -{3\over 2}$. Now take any point $T(x_0,y_0)$ from one line, say $d_1:\;\; -3x+4y+4=0$ and calcluate it distance from $d_2$, say by formula: $$ {|ax_0+by_0+c|\over \sqrt{a^2+b^2}}$$
