The Foot of the $⊥$ from origin to the plane $3 x + 4 y - 6 z + 1 = 0$ is

- \frac{3}{61}, \frac{4}{61}, \frac{6}{61}

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\frac{- 3}{61}, \frac{- 4}{61}, \frac{- 6}{61}

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\frac{4}{61}, \frac{- 3}{61}, \frac{5}{61}

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None of these

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Solution

The correct option is D None of these
Clearly direction ratios of perpendicular drawn from origin to the given plane are $3, 4, - 6$
Hence equation of perpendicular line to the given plane and passing through origin is given by,
$\frac{x}{3} = \frac{y}{4} = \frac{z}{- 6} = k$ (say)
Now let foot of perpendicular be $P (3 k, 4 k, - 6 k)$
Also this point lie in the given plane $\Rightarrow 3 (3 k) + 4 (4 k) - 6 (- 6 k) + 1 = 0 \Rightarrow k = - \frac{1}{61}$
Hence $P \equiv (- \frac{3}{61}, - \frac{4}{61}, \frac{6}{61})$

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