The equation of the plane passing through the point ( 1 , 1 , 1 ) and through the line of intersection of x + 2 y − z + 1 = 0 and 3 x − y − 4 z + 3 = 0 is

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The equation of the plane passing through the line of intersection of the given planes is

$(x + 2 y - z + 1) + λ (3 x - y - 4 z + 3) = 0$

Since the plane passes through the point $(1, 1, 1)$ , we substitute $x = 1, y = 1, z = 1$ into the equation:

$(1 + 2 (1) - 1 + 1) + λ (3 (1) - 1 - 4 (1) + 3) = 0$

$3 + λ (1) = 0$

$λ = - 3$

Substituting $λ = - 3$ back into the equation of the plane:

$(x + 2 y - z + 1) - 3 (3 x - y - 4 z + 3) = 0$

$x + 2 y - z + 1 - 9 x + 3 y + 12 z - 9 = 0$

$- 8 x + 5 y + 11 z - 8 = 0$

$8 x - 5 y - 11 z + 8 = 0$

Answer: $8 x - 5 y - 11 z + 8 = 0$

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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