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The centroid of the triangle formed by (0, 0, 0) and the point of intersection of $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 1}{1} w i t h x = 0 a n d y = 0$ is

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$(1, 1, 1)$

$(\frac{1}{6}, \frac{- 1}{3}, \frac{1}{6})$

$(\frac{- 1}{6}, \frac{1}{3}, \frac{- 1}{6})$

$(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$

answer is B.

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Detailed Solution

Any point on the given line $(K + 1, 2 K + 1, K + 1)$

$but x = 0 \Rightarrow A (0, - 1, 0)$

$y = 0 \Rightarrow B (\frac{1}{2}, 0, \frac{1}{2}); 0 (0, 0, 0)$

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