The equation of a plane passing through the line of intersection of t…

The equation of  a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x - y + z = 3 is 

$x+ 2y + 3z - 2) + λ(x- y + z - 3) = 0 $

$ x(λ + 1) +y ( 2 - λ) + z ( 3 + λ ) - 2 - 3λ = 0 $

It is a distance of $ 2/\sqrt{3}$ units from the point (3, 1, -1).

$∴\frac{|3(λ+1)+(2 -λ) - ( 3 + λ)-2 - 3λ|}{\sqrt{(λ+1)^2+(2-λ)^2 + ( 3 +λ)^2}}=\frac{2}{\sqrt{3}}$

$⇒ \frac{|-2λ|}{\sqrt{3λ^2+4λ+14}}=\frac{2}{\sqrt{3}}$

$⇒ 3λ^2 + 4λ + 14 = 3λ^2 ⇒ 4λ = - 14 ⇒ λ =\frac{-7}{2}$

Putting $λ = -\frac{7}{2}$ in (i), we obtain that the equation of the required plane is 

$-\frac{5}{2}x+\frac{11}{2}y - \frac{z}{2}+\frac{17}{2}= 0 $ or , $5x - 11y + z - 17 = 0 $