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

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

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

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

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

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Solution

The correct option is C $8 x - 5 y - 11 z + 8 = 0$
Given: $p_{1} : x + 2 y - z + 1 = 0$ and
$p_{2} : 3 x - y - 4 z + 3 = 0$
Any plane through the line of intersection is
$p_{1} + λ p_{2} = 0$
$x + 2 y - z + 1 + λ (3 x - y - 4 z + 3) = 0$
It should pass through $(1, 1, 1)$
$∴ (1 + 2 - 1 + 1) + λ (3 - 1 - 4 + 3) = 0$
$\Rightarrow λ = - 3$
Hence, the required plane is: $8 x - 5 y - 11 z + 8 = 0$

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