(c)

$\because p, q, r \in Q$ and $\sqrt{p}+\sqrt{q}+\sqrt{r} \in Q$

$\Rightarrow \quad(\sqrt{p}+\sqrt{q}+\sqrt{r})^2 \in Q$

$\Rightarrow \sqrt{p q}+\sqrt{q r}+\sqrt{r} p \in Q$

Case $I$ Let exactly one of $\sqrt{p}, \sqrt{q}, \sqrt{r}$ is irrational.

$\sqrt{p}, \notin Q$ but $\sqrt{r}, \sqrt{q} \in Q$

From Eq.$(i)$, $\sqrt{p}(\underbrace{\sqrt{q}+\sqrt{r}}_{\text {rational }}) \in Q$ which is contradiction.

Case $II$ Let exactly two out of $\sqrt{p}, \sqrt{q}, \sqrt{r}$ are irrational.

$\sqrt{p}, \sqrt{q} \notin Q$ but $\sqrt{r} \in Q$

From Eq. $(i)$,

$\begin{aligned} & \sqrt{p} \sqrt{q}+\sqrt{r} \sqrt{p}+\sqrt{r} \sqrt{q}+(\sqrt{r})^2 \in Q \\ \Rightarrow &(\sqrt{p}+\sqrt{r})(\sqrt{q}+\sqrt{r}) \in Q \end{aligned}$

$\because$ Both $\sqrt{p}+\sqrt{r}$ and $\sqrt{q}+\sqrt{r}$ are irrational.

Hence, they must be conjugate of each other. Which is contradiction.

Case $III$ Let all $\sqrt{p}, \sqrt{q}, \sqrt{r}$ are irrational.

Let $\sqrt{p}+\sqrt{q}+\sqrt{r}=x$, when $x \in Q^{+}$

From Eq. (i), $\sqrt{p}(x-\sqrt{p})+\sqrt{q} \sqrt{r} \in Q$

$\Rightarrow x \sqrt{p}+\sqrt{q} \sqrt{r} \in Q$

Which is contradiction.

Hence, all $\sqrt{p}, \sqrt{q}$ and $\sqrt{r}$ must be rational.