If the equation \(7 x^2-14 x y+p y^2-12 x+q y-4=0\) represents a pair of parallel lines then the value of \(\sqrt{p^2+q^2-p q}\) is

Given equation of pair of lines is
\(7 x^2-14 x y+p y^2-12 x+q y-4=0 \)
\( \therefore \mathrm{a}=7, \mathrm{~b}=\mathrm{p}, \mathrm{c}=-4, \mathrm{~h}=-7, \mathrm{~g}=-6, \mathrm{f}=\frac{\mathrm{q}}{2}\)
The lines are parallel.
\(\therefore h^2=a b \)
\( \Rightarrow(-7)^2=7 p \)
\( \Rightarrow p=7\)
Now, \(\mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^2-\mathrm{bg}^2-\mathrm{ch}^2=0\)
\(\Rightarrow 7(7)(-4)+2\left(\frac{q}{2}\right) (-6)(-7) \) \( -7\left(\frac{\mathrm{q}}{2}\right)^2-7(-6)^2-(-4)(-7)^2=0\)
\(\Rightarrow-196+42 q-\frac{7 q^2}{4}-252+196=0 \)
\( \Rightarrow q^2-24 q+144=0 \)
\( \Rightarrow(q-12)^2=0 \)
\( \Rightarrow q=12 \)
\( \therefore \sqrt{p^2+q^2-p q}=\sqrt{49+144-84}=\sqrt{109}\)