be the equation of line of intersection of plans: 3x - 6y - 2z = 15 2x + y - 2z = 5 Since the line of intersection lies in both planes, we get 3a - 6 - 2c = 0 2a + b - 2c = 0 That is,
a
14
=
b
2
=
c
15
⇒ a = 14 ; b = 2 ; c = 15 Let (x1,y1,0) be a point that lies on the line of intersection; therefore, 3x1−6y1 = 15 2x1+y1 = 5 ⇒ (x1,y1,z1) = (3 , - 1 , 0) Hence, the equation of line of intersection is
x−3
14
=
y+1
2
=
z
15
= t That is, x = 3 + 14t y = 2t - 1 z = 15 t Therefore, Statement-1 is false and Statement-2 is true.