Let $\overrightarrow{a}, \overrightarrow{b} and \overrightarrow{c}$ be three non-coplanar vectors and $\overrightarrow{d}$ be a non-zero vector, which is perpendicular to $\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$ . Now $\overrightarrow{d}=(\overrightarrow{a}\times \overrightarrow{b})sinx+(\overrightarrow{b}\times \overrightarrow{c})cosy+2(\overrightarrow{c}\times \overrightarrow{a})$. Then

$\overrightarrow{d}.\overrightarrow{a}=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]cosy=-\overrightarrow{d}.(\overrightarrow{b}+\overrightarrow{c})$

$orcosy=-\frac{\overrightarrow{d}.(\overrightarrow{b}+\overrightarrow{c})}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$

Similarly, $sinx=\frac{\overrightarrow{d}.(\overrightarrow{a}+\overrightarrow{b})}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}and\frac{\overrightarrow{d}.(\overrightarrow{a}+\overrightarrow{c})}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}=-2$

$\therefore sinx+cosy+2=0$

$orsinx+cosy=-2$

$orsinx=-1,cosy=-1$

Since we want the minimum value of $x^2+y^2,x=-\pi /2,y=\pi$

Therefore, the minimum value or $x^2+y^2 is 5{\pi }^2/4$