$p_1^2 + p_2^2 = q_1^2 + q_2^2 = r_1^2 + r_2^2 = 2$, $p_1^2 + q_1^2 + r_1^2 = 3$, positive distinct rational solution I would like to know if equations

$p_1^2 + p_2^2 = 2$

$q_1^2 + q_2^2 = 2$

$r_1^2 + r_2^2 = 2$

$p_1^2 + q_1^2 + r_1^2 = 3$

have a solution where $p_1,p_2,q_1,q_2,r_1,r_2$ are pairwise distinct and positive rational numbers.

I will appreciate any help.
Yes, the possible solutions in the rationals are $x,y = \pm1, \pm1$, so

$$p = [1,1]$$
$$q = [-1,1]$$
$$r = [-1,-1]$$

is a solution

Edit:

Based on comments.

We seek a solution $x,y$ such that $x^2 +y^2=2$ where $x,y \in \mathbb{Q}$. To fulfill this (and without loss of generality within the specified domain) we require that $0 \le x \le 1$ and $1\le y\le 2$. Let's then define two new variables $a,b$ and require $0 \le a,b \le 1$ and reexpress our equation as

$$(1+a)^2 + (1-b)^2 = 2$$

Expanding this equation

$$ a^2 + 2 a + 1 + b^2 - 2b + 1 = 2$$

Simplifying,

$$ b^2 - 2b + a^2 + 2 a= 0$$

Therefore,

$$ a^2 + b^2 = 2 ( b - a )$$

so $b \ge a$

If $b = a + \epsilon$, for some non-negative $\epsilon$

$$ a^2 + (a+\epsilon)^2 = 2 ( a+\epsilon - a )$$

$$ 2 a^2 + 2a\epsilon + \epsilon^2 = 2 \epsilon $$

Solving for $a$

$$a = \frac{-2 \epsilon \pm \sqrt{(2 \epsilon)^2 - 4 (2) (\epsilon^2 - 2 \epsilon ) }}{2 \cdot 2}$$

$$a = \frac{-\epsilon \pm \sqrt{4 \epsilon - \epsilon^2}}{2}$$

so for $a$ to be real, $0 \le \epsilon \le 4$. And for $a$ to be rational, there must be some coprime integers $m,n$ such that

$$4 \epsilon - \epsilon^2 = \left( \frac{m}{n} \right)^2$$

Or equivalently

$$ n^2( 4 \epsilon - \epsilon^2 ) = m^2$$

The only way for this to be the case is if $4 \epsilon - \epsilon^2$ is a square number, which within the domain limits on $\epsilon$ means $\epsilon=0,4$

This corresponds to $a=0$ and $b=0$ or $b=2$, thus the solutions are $x,y= 1,\pm1$ and cannot be distinct and positive.

Note:

Because of the definition of $a$, we lose the $[-1,-1]$ solution
Yes, the possible solutions in the rationals are $x,y = \pm1, \pm1$, so

$$p = [1,1]$$
$$q = [-1,1]$$
$$r = [-1,-1]$$

is a solution

Edit:

Based on comments.

We seek a solution $x,y$ such that $x^2 +y^2=2$ where $x,y \in \mathbb{Q}$. To fulfill this (and without loss of generality within the specified domain) we require that $0 \le x \le 1$ and $1\le y\le 2$. Let's then define two new variables $a,b$ and require $0 \le a,b \le 1$ and reexpress our equation as

$$(1+a)^2 + (1-b)^2 = 2$$

Expanding this equation

$$ a^2 + 2 a + 1 + b^2 - 2b + 1 = 2$$

Simplifying,

$$ b^2 - 2b + a^2 + 2 a= 0$$

Therefore,

$$ a^2 + b^2 = 2 ( b - a )$$

so $b \ge a$

If $b = a + \epsilon$, for some non-negative $\epsilon$

$$ a^2 + (a+\epsilon)^2 = 2 ( a+\epsilon - a )$$

$$ 2 a^2 + 2a\epsilon + \epsilon^2 = 2 \epsilon $$

Solving for $a$

$$a = \frac{-2 \epsilon \pm \sqrt{(2 \epsilon)^2 - 4 (2) (\epsilon^2 - 2 \epsilon ) }}{2 \cdot 2}$$

$$a = \frac{-\epsilon \pm \sqrt{4 \epsilon - \epsilon^2}}{2}$$

so for $a$ to be real, $0 \le \epsilon \le 4$. And for $a$ to be rational, there must be some coprime integers $m,n$ such that

$$4 \epsilon - \epsilon^2 = \left( \frac{m}{n} \right)^2$$

Or equivalently

$$ n^2( 4 \epsilon - \epsilon^2 ) = m^2$$

The only way for this to be the case is if $4 \epsilon - \epsilon^2$ is a square number, which within the domain limits on $\epsilon$ means $\epsilon=0,4$

This corresponds to $a=0$ and $b=0$ or $b=2$, thus the solutions are $x,y= 1,\pm1$ and cannot be distinct and positive.

Note:

Because of the definition of $a$, we lose the $[-1,-1]$ solution