Find the equation of the plane through the line of intersection of the plane x + y + z = 1 and 2x + 3y + 4z = 5 and twice of its y–intercept is equals to the three times its z intercept?

Question

Philoid · Accepted Answer

We know that equation of plane passing through the line of intersection of planes

(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

So equation of plane passing 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

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

so,
as given that twice of its y intercept is equals to the three times its z intercept

so

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

2 + 8k = 3 + 9k

k = – 1

put this in equation (1)

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

x[1 + 2( – 1)] + y[1 + 3( – 1)] + z[1 + 4( – 1)] – 1 – 5( – 1) = 0

– x – 2y – 3z + 4 = 0

x + 2y + 3z = 4