A sequence includes $a_p=\sqrt2$, $a_q=\sqrt3$, $a_r=\sqrt5$ for some $1\leq p
In a sequence $a_1, a_2,\dots$ of real numbers it is observed that $a_p=\sqrt{2}$, $a_q=\sqrt{3}$, and $a_r=\sqrt{5}$, where $1\leq p<q<r$ are positive integers. Then $a_p$, $a_q$, $a_r$ can be terms of
(A) an arithmetic progression
(B) a harmonic progression
(C) an arithmetic progression if and only if $p$, $q$, and $r$ are perfect squares
(D) neither an arithmetic progression nor a harmonic progression
I have tried using the definition of AP and argued that if the first option is true then $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{q-p}{r-q}$  is true which implies that the left hand side is an rational number. I don't know whether the left hand side is a rational number or not. I am stuck here. Also, I don't know how correct this approach is. How do I approach and solve this problem?

Suppose that $x=\sqrt{15}-\sqrt{10}-\sqrt{6}$ were rational.
Then, squaring the number would give the following:
$$\begin{align}
x^2
&=(\sqrt{15}-\sqrt{10}-\sqrt{6})(\sqrt{15}-\sqrt{10}-\sqrt{6})\\
&=15-\sqrt{150}-\sqrt{90}-\sqrt{150}+10+\sqrt{60}-\sqrt{90}+\sqrt{60}+6\\
&=31-2\sqrt{150}-2\sqrt{90}+2\sqrt{60}\\
&=31-2(5\sqrt{6})-2(3\sqrt{10})+2(2\sqrt{15})\\
&=31-10\sqrt{6}-6\sqrt{10}+4\sqrt{15}
\end{align}$$
Next, to get rid of the $\sqrt{6}$ term, one must multiply $x$ by $-10$ to cancel the $\sqrt{6}$ term from $x^2$. This gives $x^2-10x=(31-10\sqrt{6}-6\sqrt{10}+4\sqrt{15})+(-10\sqrt{15}+10\sqrt{10}+10\sqrt{6})=31+4\sqrt{10}-6\sqrt{15}$.
The above would then lead to the rationality of $31+4\sqrt{10}-6\sqrt{15}$. Since $31$ is clearly rational, we may ignore that term, leading to the rationality of $x^2-10x-31=4\sqrt{10}-6\sqrt{15}$.
Squaring one more time, we get the rationality of $(x^2-10x-31)^2=(4\sqrt{10}-6\sqrt{15})^2=160-48\sqrt{150}+540=700-48(5\sqrt{6})=700-240\sqrt{6}$.
Finally, subtracting $700$ and then dividing by $-240$ leads to the rationality of $\sqrt{6}$.
But we know that square roots of integers are either themselves integers or else irrational, leading to a contradiction. Hence, $x=\sqrt{15}-\sqrt{10}-\sqrt{6}$ must be irrational.
Alternatively, one could instead get rid of the $\sqrt{10}$ term by considering $x^2-6x$, or the $\sqrt{15}$ term by considering $x^2-4x$, and similar steps would show that the original term that one got rid of is rational.
