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Plane through intersection, perpendicular to plane

Short answer

Plane through intersection: (7x-4y+7z+16)+λ(4x-3y-2z+13)=0. Normal (7+4λ, -4-3λ, 7-2λ) dot (1,-1,-2) = 0 gives λ=3/11. Substituting yields 89x - 53y + 71z + 215 = 0.

Obtain the equation of the plane passing through the line of intersection of two planes 7x−4y+7z+16 = 0 and 4x−3y−2z+13 = 0 and perpendicular to plane x−y−2z+5 = 0.

Given planes. P₁: 7x − 4y + 7z + 16 = 0 and P₂: 4x − 3y − 2z + 13 = 0.

Plane through their line of intersection

(7x − 4y + 7z + 16) + λ(4x − 3y − 2z + 13) = 0

(7 + 4λ)x − (4 + 3λ)y + (7 − 2λ)z + (16 + 13λ) = 0     ... (1)

Its normal is (7 + 4λ, −(4 + 3λ), 7 − 2λ).

Perpendicular to x − y − 2z + 5 = 0

Normals must be perpendicular: their dot product is zero. The given plane has normal (1, −1, −2).

(7 + 4λ)(1) + (−(4 + 3λ))(−1) + (7 − 2λ)(−2) = 0
7 + 4λ + 4 + 3λ − 14 + 4λ = 0
11λ − 3 = 0   →   λ = 3/11

Substitute in (1)

Multiply (1) by 11 and use λ = 3/11:

(77 + 12)x − (44 + 9)y + (77 − 6)z + (176 + 39) = 0
89x − 53y + 71z + 215 = 0

Check: the normal (89, −53, 71) with (1, −1, −2): 89 + 53 − 142 = 0 ✓.

Answer: 89x − 53y + 71z + 215 = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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