If a plane has X-intercept $l$ , Y-intercept $m$ and Z-intercept $n$ , and perpendicular distance of plane from origin is $k$ , then ?

l^{2} + m^{2} + n^{2} = \frac{1}{k^{2}}

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\frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}} = \frac{1}{k^{2}}

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l^{2} + m^{2} + n^{2} = k^{2}

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\frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}} = k^{2}

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Solution

The correct option is D $\frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}} = \frac{1}{k^{2}}$
The equation of plane is
$\frac{x}{l} + \frac{y}{m} + \frac{z}{n} = 1$

$⟹ \frac{x}{l} + \frac{y}{m} + \frac{z}{n} - 1 = 0$

Distance from origin $= \frac{| \frac{0}{l} + \frac{0}{m} + \frac{0}{n} - 1 |}{\sqrt{(\frac{1}{l})^{2} + (\frac{1}{m})^{2} + (\frac{1}{n})^{2}}}$

$⟹ k = \frac{1}{\sqrt{(\frac{1}{l})^{2} + (\frac{1}{m})^{2} + (\frac{1}{n})^{2}}}$

$⟹ \frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}} = \frac{1}{k^{2}}$

Hence, the answer is option (B).

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