Here, we have to find the equation of the plane passing through the points A (3, 1, 2), B (6, 1, 2) and C (0, 2, 0)
Here,
x1=3,y1=1,z1=2,x2=6,y2=1, z2=2,x3=0,y3=2 and
z3=0 As we know that, equation of the plane in Cartesian form passing through three non collinear points
(x1,y1,z1),(x2,y2,z2) and
(x3,y3,z3) is given by:
|| x−x1 | y−y1 | z−z1 |
| x2−x1 | y2−y1 | z2−z1 |
| x3−x1 | y3−y1 | z3−z1 |
| =0 ⇒|| =0 ⇒(x−3)×(0−0)−(y−1)×(−6−0) +(z−2)×(3−0)=0 ⇒6y−6+3z−6=0 ⇒6y+3z−12=0 So, the equation of the required plane is 6y + 3z - 12 = 0
Hence, option D is the correct answer.