A triangular plane $A B C$ with centroid $(1, 2, 3)$ cuts the coordinates axes at $A, B, C$ respectively.
What is the equation of the plane $A B C$ ?

x + 2 y + 3 z = 1

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3 x + 2 y + z = 3

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2 x + 3 y + 6 z = 18

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6 x + 3 y + 2 z = 18

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Solution

The correct option is D $6 x + 3 y + 2 z = 18$
$A = (x, 0, 0), B = (0, y, 0), C = (0, 0, z)$
Centroid of the triangle ABC = $(\frac{x + 0 + 0}{3}, \frac{0 + y + 0}{3}, \frac{0 + 0 + z}{3})$
Hence $(\frac{x}{3}, \frac{y}{3}, \frac{z}{3}) = (1, 2, 3)$
$∴$ The intercepts made by the plane $A B C$ on the axes are $(x, y, z) = (3, 6, 9)$
Hence the equation of the plane $A B C$ is $\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$
$\Rightarrow 6 x + 3 y + 2 z = 18$

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The equation of the plane passing through the origin and perpendicular to the line x=2 y=3 z is

\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0

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