If $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$ $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$ and $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$ $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$ where a, b, c, p, q, r are non-zero real numbers, then $\frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}}$ $\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}$ is equal to = ?

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0 \\ \Rightarrow a q r + b p r + c p q = 0.... (i) \\ \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1 \\ \Rightarrow {(\frac{p}{a} + \frac{q}{b} + \frac{r}{c})}^{2} = 1 \\ \Rightarrow \frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}} + 2 (\frac{p q}{a b} + \frac{p r}{a c} + \frac{q r}{b c}) = 1 \\ \Rightarrow \frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}} + \frac{2 (p q c + p r b + q r a)}{a b c} = 1 \\ \Rightarrow \frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}} = 1.... [using (i)] \end{aligned}

$\eqalign{ & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0\,\,\,\, \cr & \Rightarrow aqr + bpr + cpq = 0....({\text{i}}) \cr & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1\,\, \cr & \Rightarrow {\left( {\frac{p}{a} + \frac{q}{b} + \frac{r}{c}} \right)^2} = 1 \cr & \Rightarrow {\text{ }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}\, + 2\left( {\frac{{pq}}{{ab}} + \frac{{pr}}{{ac}} + \frac{{qr}}{{bc}}} \right) = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} + \frac{{2\left( {pqc + prb + qra} \right)}}{{abc}} = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1....\left[ {{\text{using (i)}}} \right] \cr}$