Find the equation of plane passing through the points A (3, 2, 1), B (4, 2, - 2) and C(6, 5, - 1) and hence find the value of 
for which A (3, 2, 1), B(4, 2, - 2), C(6, 5, - 1) and 
are coplanar
OR
Find the co - ordinates of the point where the line 
meets the plane which is perpendicular to the vector 
and at a distance of 
from origin[CBSE 2016]
Given: A (3, 2, 1) B (4, 2, - 2) C (6, 5, - 1)
Formula: equation of plane is given by
Substituting the values in above equation
(x - 3)9 – (y - 2)7 + (z - 1)3 = 0
9x - 7y + 3z = 16
Now if A, B, C, and D are co - planar D must lie on the plane
9λ - 35 + 15 - 16 = 0
λ = 4
OR
given: 
Equation of plane perpendicular to 
at a distance of 
from origin is
= (-1 + 3λ)1 + (-2 + 4λ)1 + (-3 + 3λ)3 = 4
16λ = 16
λ = 1
Hence substituting λ = 1 in 
The point of intersection is ( - 1 + 3, - 2 + 2, - 3 + 3) which is
(2, 2, 0)
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.
[CBSE 2011]