The parametric form of the equation of the line 3x-6y-2z-15=0=2x+y-2z-5 is

The given planes are 3x-6y-2z-15=0=2x+y-2z-5To get a point on the line of intersection of the above two planes, substitute z=0it gives 3x-6y=15,2x+y=5solving the above two equations, we get P3,-1,0Suppose that the direction ratios of line are ⟨a,b,c⟩Hence, 3a−6b−2c=0,2a+b−2c=0 it implies that a14=b2=c15     by cross multiplication methodTherefore, the equation of the line x−314=y+12=z15