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771.
A plane P meets the coordinate axes at A, B and C respectively. The centroid of $∆$ ABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is:

$\frac{x - 1}{2} = \frac{y - 1}{1} = \frac{z - 2}{1}$

$\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$

$\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 2}{2}$

$\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 2}{1}$

$\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 2}{1}$

$Let A (α, 0, 0), B (0, β, 0), C (0, 0, γ) thenG (\frac{α}{3}, \frac{β}{3}, \frac{γ}{3}) \equiv (1, 1, 2)$

$α = 3, β = 3, γ = 6 ∴ equation of plane is \frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 1 \Rightarrow \frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1 \Rightarrow 2 x + 2 y + z = 6 ∴ required line \frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 2}{1}$

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