$\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$

$\frac{{x + p + x + q}}{{(x + p)(x + q)}} = \frac{1}{r}$

$(2x + p + q)r = {x^2} + px + qx + pq$

${x^2} + (p + q – 2r)x + pq – pr – qr = 0$

Let $\alpha$ and $\beta$ be the roots.

$\therefore \alpha+\beta=-(p+q-2 r)……….(i)$

$\alpha \beta=pq-pr-qr……….(ii)$

$\because \alpha=-\beta(\text { given })$

$\therefore$ in eq. $(1),$ we get

$\Rightarrow \quad-(p+q-2 r)=0……….(iii)$

Now, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$

$=(-(p+q-2 r))^{2}-2(p q-p r-q r) \ldots$ (from $(i)$ and $(ii))$

$=p^{2}+q^{2}+4 r^{2}+2 p q-4 p r-4 q r-2 p q+2 p r+2 q r$

$=p^{2}+q^{2}+4 r^{2}-2 p r-2 q r$

$=p^{2}+q^{2}+2 r(2 r-p-q) \ldots(\text { from (iii) })$

$=p^{2}+q^{2}+0$

$=p^{2}+q^{2}$