### Coordinates and vectors

Let $\vec{r}_0$ be the center of a sphere and $R$ be its radius. We have
$(\vec{r}_i-\vec{r}_0)^2=R^2$
where $i=1,2,3,4$. Subtracting, say, equation for $i=1$ from others, we have
$\vec{r}_0\cdot \vec{X}_2=\vec{r}_2^2-\vec{r}_1^2\\ \vec{r}_0\cdot \vec{X}_3=\vec{r}_3^2-\vec{r}_1^2\\\vec{r}_0\cdot \vec{X}_4=\vec{r}_4^2-\vec{r}_1^2$
where $\vec{X}_j=\vec{r}_j-\vec{r}_1$, $j=2,3,4$. We assume that $\vec{X}_j$ are not all in the same plane. Then we can decompose $\vec{r}_0=k_2\vec{X}_2+k_3\vec{K_3}+k_4\vec{K}_4$ and substituting one has
$k_2\vec{X}_2\cdot\vec{K}_2 + k_3\vec{X_3}\cdot\vec{X}_2+ k_4\vec{X_4}\cdot\vec{X}_2=\vec{r}_2^2-\vec{r}_1^2\\ k_2\vec{X}_2\cdot\vec{K}_3 + k_3\vec{X_3}\cdot\vec{X}_3+ k_4\vec{X_4}\cdot\vec{X}_3=\vec{r}_3^2-\vec{r}_1^2\\ k_2\vec{X}_2\cdot\vec{K}_4 + k_3\vec{X_3}\cdot\vec{X}_4+ k_4\vec{X_4}\cdot\vec{X}_4=\vec{r}_4^2-\vec{r}_1^2$
These equations always have a solution (why ?).