Very easy problem. There are so many parallels here.

can i know the answer? i solved it, but i think my way is too hard and hard to write

How about using vectors. It can be shown that if the vertices of a tetrahedron are at $\mathbf{0,a,b,c}$, then the circumcenter is at: \[\frac{(\mathbf{a\cdot a})(\mathbf{b\times c})+(\mathbf{b\cdot b})(\mathbf{c\times a})+(\mathbf{c\cdot c})(\mathbf{a\times b})}{12V}\] where $V$ is the volume. Now WLOG, assume that $A_2=-aA_1$, $B_2=-bB_1$, and $C_2=-cC_1$, where $a,b,c\in\mathbb{R}^+$ so that $P$ is at the origin. Then we have: $O_{111}:\ \frac{(A_1\cdot A_1)(B_1\times C_1)+(B_1\cdot B_1)(C_1\times A_1)+(C_1\cdot C_1)(A_1\times B_1)}{12V_{111}} $ $O_{222}:\ \frac{-a(A_1\cdot A_1)(B_1\times C_1)-b(B_1\cdot B_1)(C_1\times A_1)-c(C_1\cdot C_1)(A_1\times B_1)}{12V_{111}}$ Similarly for the other circumcenters. From this we see that the circumcenters are vertices of a parallelepiped and the four segments asked for are the body diagonals of that parallelepiped, so they are concurrent and even happen to bisect each other as well.

What a troll problem Lemma: the locus of points $O$ in space such that $OA=OB=OC$ for three points $A, B, C$ is the line passing through the circumcenter of $\triangle ABC$ perpendicular to the plane containing $\triangle ABC$. Proof: the locus is the intersection of the planes perpendicular to the segments $AB, BC, CA$ passing through their midpoints, which is exactly the line described. There does not remain much to be done. Since $A_1A_2\cap B_1B_2\ne \emptyset$, $A_1, A_2, B_1, B_2$ lie on a plane $\mathcal{P}_{12}$. Then by the lemma $O_{ij1}O_{ij2}\perp \mathcal{P}_{12}$ for all $i,j\in\{1,2\}$, so $O_{ij1}O_{ij2}\| O_{kl1}O_{kl2}$ for $i,j,k,l \in\{1,2\}$. This is true for planes $\mathcal{P}_{23}$ and $\mathcal{P}_{13}$ so in fact the solid created by all the circumcenters is just a rectangular prism. But we're done, since the space diagonals of a rectangular prism obviously intersect the center.

$\textbf{Proof :}$ Consider inversion wrt sphere with center at $P$. Let $A_1', A_2',\ldots , C_2'$ be images of points $A_1, A_2,\ldots , C_2$ respectively. Let spheres $(O_{ijk}), (O_{\bar{i}\bar{j}\bar{k}})$ intersect by circles $C_{ijk}$. To prove that lines $O_{ijk}O_{\bar{i}\bar{j}\bar{k}}$ are concurrent it's enough to prove that circles $C_{ijk}$ lie on the same sphere. Consider intersection line $L_{ijk}$ of planes $<A_iB_jC_k>$ and $<A_{\bar{i}}B_{\bar{j}}C_{\bar{k}}>$. Line $L_{ijk}$ is image of circle $C_{ijk}$, so it's enough to prove that all lines $L_{ijk}$ lie on the same plane. For any $i, j, k, p$ easy to see that lines $L_{ijk}$, $L_{ijp}$ intersect at point $A_iB_j\cap A_{\bar{i}}B_{\bar{j}}$. So any two line $L_{ijk}$, $L_{ijp}$ lie on the same plane. So all lines $L_{ijk}$ lie on the same plane. $\Box$

@bobthesmartypants Actually, circumcenters create parallelepiped but solutions still works