Correct Answer - Option 3 : 3x - 3y + z = 10
Concept used:
The equation of the plane passing through (x1, y1, z1) is
a(x − x1) + b(y − y1) + c(z − z1) = 0
Here, a, b and c are direction ratios
If the foot of the perpendicular drawn from (-2, 1, 0), Then
Direction ratios of normal to the required plane are given by
a = (x1 - (-2)), b = (y1 - 1) and c = ( z1 - 0)
Where (x1, y1, z1) is Point of the plane
Calculation:
(x1, y1, z1) is (1, -2, 1)
Direction ratios of normal to the required plane are given by
a = (x1 - (-2)), b = (y1 - 1) and c = ( z1 - 0)
⇒ a = 3, b = - 3 and c = 1
The equation of the plane passing through (x1, y1, z1) is
a(x − x1) + b(y − y1) + c(z − z1) = 0
⇒ 3(x - 1) - 3(y + 2) + 1(z - 1) = 0
⇒ 3x - 3 - 3y - 6 + z - 1 = 0
∴ The equation of the plane is 3x - 3y + z = 10
