Question

Find the equation of the plane passing through the intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and the point (1, 1, 1).​

Answer 1

Formula :

Let two planes be

  U : ax + by + cz + d = 0

  V : ex + fy + gz + h = 0

Then, the equation of the plane passing through U and V planes be

  U + kV = 0 , where k is to determined from other given conditions.

Solution of the problem :

The given planes are

    x + y + z - 6 = 0

    2x + 3y + 4z + 5 = 0

Then, the place passing through their intersection can be taken as

(x + y + z - 6) + k (2x + 3y + 4z + 5) = 0 ...(i)

Given that, (i) no. plane passes through the point (1, 1, 1) and thus (1, 1, 1) satisfies (i) no. equation. Then -

  (1 + 1 + 1 - 6) + k (2 + 3 + 4 + 5) = 0

  or, - 3 + 14k = 0

  or, k = 3/14

Hence, the required plane is

(x + y + z - 6) + (3/14) (2x + 3y + 4z + 5) = 0

or, 14 (x + y + z - 6) + 3 (2x + 3y + 4z + 5) = 0

or, 14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0

or, 20x + 23y + 26z - 69 = 0

         or, 20x + 23y + 26z = 69

Answer 2

here is your answer

PLZ refer to the attachment for answer..

thanks