Prove that if a plane has the intercepts a, b, c and if it is at a distance of p units from the origin, then $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$

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Solution

The equation of a plane having intercepts a, b, c, with X, Y and Z-axes respectively, is given by
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \Rightarrow (\frac{1}{a}) x + (\frac{1}{b}) y + (\frac{1}{c}) z - 1 = 0$
Distance of this plane from the origin (0, 0, 0,) is given to be p.
$∴ p = \frac{∣ ∣ \frac{1}{a} .0 + \frac{1}{b} .0 + \frac{1}{c} .0 - 1 ∣ ∣}{\sqrt{{(\frac{1}{a})}^{2} + {(\frac{1}{b})}^{2} + {(\frac{1}{c})}^{2}}} \Rightarrow \frac{1}{p} = \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{} c^{2}}$
On squaring both sides, we get $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}$

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Prove that if a plane has the intercepts a, b, c and if it is at a distance of p units from the origin, then 1a2+1b2+1c2=1p2

Prove that if a plane has the intercepts a, b, c and if it is at a distance of p units from the origin, then $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$