If $\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1$ and $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$ where $a, b, c, p$, $q, r$ are non-zero real numbers, then $\frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}+\frac{r^{2}}{c^{2}}$ is equal to
(a)0
(b)1
(c)3
(d)9
Answer
Answer (as printed): B
Explanation
$\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0 \Rightarrow a q r+b p r+c p q=0$ $$\begin{aligned} & \frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1 \Rightarrow\left(\frac{p}{a}+\frac{q}{b}+\frac{r}{c}\right)^{2}=1 \\ & \Rightarrow \frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}+\frac{r^{2}}{c^{2}}+2\left(\frac{p q}{a b}+\frac{p r}{a c}+\frac{q r}{b c}\right)=1 \\ & \Rightarrow \frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}+\frac{r^{2}}{c^{2}}+\frac{2(p q c+p r b+q r a)}{a b c}=1 \\ & \Rightarrow \frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}+\frac{r^{2}}{c^{2}}=1 \end{aligned}$$ [Using (i)]
Explanation as extracted from the printed page; notation may be imperfect.