The equation $8x^2-24xy+18y^2-6x+9y-5=0$ represents a

We know that a general equation of second degree

$a{x}^2+ 2hxy+ b{y}^2 + 2gx + 2fy + c = 0 ...\left(1\right)$

represents a pair of straight line if

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$ and $h^2>0$,

And $\left(1\right)$ represents a pair of parallel lines, if 

$f·g\ne 0, b{g}^2=a{f}^2=k\ge abc \mathrm{for} a>0 \mathrm{and} k\le abc \mathrm{for} a<0.$

We have the equation

$8x^2-24xy+18y^2-6x+9y-5=0 ...\left(2\right)$
Here, the value $\left|\begin{array}{ccc}8& -12& -3\\ -12& 18& \frac{9}{2}\\ -3& \frac{9}{2}& -5\end{array}\right|=8\left\{18\times \left(-5\right)-\frac{9}{2}\times \frac{9}{2}\right\}+12\left\{\left(-12\right)\times \left(-5\right)+3\times \frac{9}{2}\right\}-3\left\{\left(-12\right)\times \frac{9}{2}-\left(-3\right)\times 18\right\}$

$=8\left(-90-\frac{81}{4}\right)+12\left(60+\frac{27}{2}\right)-3\left(-54+54\right)=0$.

Also, as,  we have

 $a=8>0$, so

 $18\times 3^2=8\times {\left(\frac{9}{2}\right)}^2$ which is $\ge 8\times 18\times \left(-5\right)$

As, equation $\left(2\right)$ satisfies all the above mentioned conditions, hence, it represents a pair of parallel lines.