The direction vector of the line of intersection of the planes x -3y +7 = 0 and x -2z - 4 = 0 is the cross product of their normal vectors.

<1 -3 0> `xx` <1 0 -2> = < 6 2 3>

The normal vector of the plane 3x - 6y - 2z -12 =0 is <3 -6 -2>

The vectors <6 2 3> and <3 -6 -2> are not parallel and as their cross product is not 0 neither are they perpendicular.

**The line of intersection of the planes x -3y +7 = 0 and x -2z - 4 = 0 is neither parallel nor perpendicular to the the plane 3x - 6y - 2z -12 =0**

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