# 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 coplanarORFind 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)

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