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Question
Find the equation of the plane through the intersection of the planes 3x − y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).
Solution
` \text{ The equation of the plane passing through the line of intersection of the given planes is }`
\[3x - y + 2z - 4 + \lambda \left( x + y + z - 2 \right) = 0 . . . \left( 1 \right)\]
`\text{ This passes through (2, 2, 1). So,} `
\[6 - 2 + 2 - 4 + \lambda \left( 2 + 2 + 1 - 2 \right) = 0\]
\[ \Rightarrow 2 + 3\lambda = 0\]
\[ \Rightarrow \lambda = \frac{- 2}{3}\]
`\text{ Substituting this in (1), we get }`
\[3x - y + 2z - 4 - \frac{2}{3} \left( x + y + z - 2 \right) = 0\]
\[ \Rightarrow 9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0\]
\[ \Rightarrow 7x - 5y + 4z = 8\]
\[\]
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