We have,
$ 6x^{2} + xy -12y^{2} -13 x +6y +6 = 0$
Here, $a=6, h= \frac{1}{2}, b = -12$,
$ g = - \frac{13}{2}$
$f = 3, c =6 $
$\because a+b = 6 -12 = -6 \ne 0 $
$∴$ A pair of line is not perpendicular and
$\because h^2 \ne ab$
$∴$ Pair of line is not parallel
Hence, pair of straight line is not passing through origin, neither parallel nor perpendicular.