The parametric equation of the line of intersection of the given planes are

$3x-6y-2z=15,$ and $2x+y-2z=5$are

detailed solution

Correct option is C

the given planes  are 3x-6y-2z=15, and 2x+y-2z=5 The line of intersection of two planes is parallel to the vector which is cross product of the normal vectors to the planes hence the vector along the line of intersection of two planes is ijk3−6−221−2=i(12+2)−j(−6+4)+k(3+12)This can be simplify as 14i-2j+15kTo get a point on the line of intersection of two planes, substitute  z=0 in the plane equations and then solve for the other two variablesit implies that 3x-6y=15,2x+y=5solving the above two simultaneous equations x=3,y=1Therefore, the parametric form of the equation of the line of intersection of two planes is x=14t+3,y=2t+1,z=15t