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

.: The equation of the plane in the intercept form is

$\cfrac{x}{a} + \cfrac{y}{b} + \cfrac{z}{c} = 1$ or $\left( {\cfrac{1}{a}} \right)x + \left( {\cfrac{1}{b}} \right)y + \left( {\cfrac{1}{c}} \right)z - 1 = 0$

Distance of the plane from origin is p units.

$\Rightarrow$ $\cfrac{{\left| {\cfrac{1}{a} \cdot (0) + \cfrac{1}{b}(0) + \cfrac{1}{c}(0) - 1} \right|}}{{\sqrt {{{\left( {\cfrac{1}{a}} \right)}^2} + {{\left( {\cfrac{1}{b}} \right)}^2} + {{\left( {\cfrac{1}{c}} \right)}^2}} }} = p \Rightarrow \cfrac{1}{p}$

$= \sqrt {\cfrac{1}{{{a^2}}} + \cfrac{1}{{{b^2}}} + \cfrac{1}{{{c^2}}}}$

$\Rightarrow$ $\cfrac{1}{{{p^2}}} = \cfrac{1}{{{a^2}}} + \cfrac{1}{{{b^2}}} + \cfrac{1}{{{c^2}}}$ (squaring both sides)

Choose the correct answer in Exercises 22 and 23.