The sum of the intercepts made by the plane $a x + b y + c z = d$ on the three axes of reference is

a + b + c

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\frac{1}{a} + \frac{1}{b} + \frac{1}{c}

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d (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})

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\frac{1}{d} \sqrt{a^{2} + b^{2} + c^{2}}

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Solution

The correct option is B $d (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
Given equation of plane is, $a x + b y + c z = d$
Assume this plane intersect axes at $A, B$ and $C$ , so coordinates are
$\Rightarrow A = (\frac{d}{a}, 0, 0), B = (0, \frac{d}{b}, 0)$ and $C = (0, 0, \frac{d}{c})$
So sum of intercepts is,
$= \frac{d}{a} + \frac{d}{b} + \frac{d}{c} = d (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
Hence, option 'C' is correct.

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