The equation of the family of planes through the line of intersection of planes

x + y + z = 1 and 2x + 3y + 4z = 5 is,

(x + y + z – 1) + k( 2x + 3y + 4z – 5) = 0 ……(1)

(2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1

It is perpendicular to the plane x – y + z = 0

(2k + 1)(1) + (3k + 1)( – 1) + (4k + 1)(1) = 5k + 1

2k + 1 – 3k – 1 + 4k + 1 = 5k + 1

K =

Sustiuting k = in eq.(1) , We get, x – z + 2 = 0 as the equation of the required plane

And its vector equation is

The equation of the family of a plane parallel to

…… (1)

If it passes through (a, b, c) then

()() = d

a + b + c = d

Substituting a + b + c = d in eq.(1), we get,

x + y + z = a + b + c as the equation of the required plane.