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If $$\frac{p}{a}$$ + $$\frac{q}{b}$$ + $$\frac{r}{c}$$ = 1 and $$\frac{a}{p}$$ + $$\frac{b}{q}$$ + $$\frac{c}{r}$$ = 0 where p, q, r and a, b, c are non - zero, then value of $$\frac{{{p^2}}}{{{a^2}}}$$ + $$\frac{{{q^2}}}{{{b^2}}}$$ + $$\frac{{{r^2}}}{{{c^2}}}$$ = ?

If $$\frac{p}{a}$$ + $$\frac{q}{b}$$ + $$\frac{r}{c}$$ = 1 and $$\frac{a}{p}$$ + $$\frac{b}{q}$$ + $$\frac{c}{r}$$ = 0 where p, q, r and a, b, c are non - zero, then value of $$\frac{{{p^2}}}{{{a^2}}}$$ + $$\frac{{{q^2}}}{{{b^2}}}$$ + $$\frac{{{r^2}}}{{{c^2}}}$$ = ? Correct Answer 1

$$\eqalign{ & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1 \cr & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0{\text{ }} \cr & \Rightarrow \frac{p}{a} = x,{\text{ }}\frac{q}{b} = y,{\text{ }}\frac{r}{c} = z \cr & \Rightarrow \left( {x + y + z} \right) = 1 \cr & {\text{Squaring the both sides}} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) = 1 \cr & {\text{and }}\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \cr & \Rightarrow \frac{{xy + yz + zx}}{{xyz}} = 0 \cr & \Rightarrow xy + yz + zx = 0 \cr & \therefore {x^2} + {y^2} + {z^2} = 1 \cr & {\text{So, }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1 \cr} $$

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