Q. 224.0( 2 Votes )

Find the equation of the plane determined by the points A(3, - 1, 2), B(5, 2, 4) and C(–1, –1, 6). Also find the distance of the point P(6, 5, 9) from the plane.[CBSE 2012, 2013]

Answer :

Given:



Expanding along R1,


(x – 3) (12 – 0) – (y + 1) (8 + 8) + (z – 2) (0 + 12) = 0


12 (x – 3) – 16 (y + 1) + 12 (z – 2) = 0


3 (x – 3) – 4 (y + 1) + 3 (z – 2) = 0


3x – 9 – 4y – 4 + 3z – 6 = 0


3x – 4y + 3z – 19 = 0 … (1)


We know that the perpendicular distance of point P from a plane is


Here a = 3, b = -4, c = 3, d = -19





Rationalizing the denominator,




The equation of the plane determined by the Given: points is 3x – 4y + 3z – 19 = 0 and the distance of Given: point P from the plane is .

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