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 ⇒|
x−3
y−1
z−2
3
0
0
−3
1
−2
|=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.