Question

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

Answer 1

The equation of any plane through the intersection of the planes,
3x-y+2z-4=0 and x+y+z-2=0, is
(3x-y+2z-4)+ \(\alpha\) (x+y+z-2)=0, where \(\alpha \in\) R...(1)

The plane passes through the point (2 ,2,1).

Therefore, this point will satisfy equation(1).

∴ (3×2-2+2×1-4)+α(2+2+1-2)=0
\(\Rightarrow \) 2+3\(\alpha\) =0

\(\Rightarrow \alpha=-\frac{2}{3}\)

Substituting \(\alpha=-\frac{2}{3}\)  in equation(1), we obtain

(3x-y+2z-4)-\(\frac{2}{3}\) (x+y+z-2)=0

\(\Rightarrow 3\) (3x-y+2z-4)-2(x+y+z-2)=0
\(\Rightarrow\) (9x-3y+6z-12)-2(x+y+z-2)=0
\(\Rightarrow \) 7x-5y+4z-8=0

This is the required equation of the plane.