Step 1: Split the pair of lines.
The equation $2x^2-3xy-2y^2=0$ is two lines through the origin. Factor it:
\[ 2x^2-3xy-2y^2=(2x+y)(x-2y)=0. \]
So the two lines are $2x+y=0$ and $x-2y=0$.
Step 2: Set up the perpendicular distance idea.
We need the perpendicular distance from $(1,2)$ to each line, then multiply the two distances.
Step 3: Distance to the first line.
For $2x+y=0$, distance $=\dfrac{|2(1)+2|}{\sqrt{2^2+1^2}}=\dfrac{4}{\sqrt5}$.
Step 4: Distance to the second line.
For $x-2y=0$, distance $=\dfrac{|1-2(2)|}{\sqrt{1^2+(-2)^2}}=\dfrac{3}{\sqrt5}$.
Step 5: Multiply the two distances.
\[ \frac{4}{\sqrt5}\cdot\frac{3}{\sqrt5}=\frac{12}{5}. \]
Step 6: Cross-check with the direct formula.
The product also equals $\dfrac{|a x_1^2+2h x_1 y_1+b y_1^2|}{\sqrt{(a-b)^2+4h^2}}=\dfrac{|2-6-8|}{\sqrt{16+9}}=\dfrac{12}{5}$. Both ways agree.
\[ \boxed{\dfrac{12}{5}} \]