BX = 2sqrt(13) = sqrt (52) = CX
And since AX is a median, OX = (1/3)12 = 4
And since triangle OBX is right
OB = sqrt (BX^2 - OX^2) = sqrt (52 - 16) = sqrt (36) = 6
sin OXB = sin OXC = OB/BX = 6 / [ 2sqrt (13)] = 3 /sqrt (13)
Note that angle OXC is obtuse....so its cosine is negative
And cos OXC = - sqrt [ 1 - sin^2 OXC ] = -sqrt [ 1 - 9/13] = -sqrt (4 /13] = -2/sqrt (13)
Using the Law of Cosines
OC^2 = OX^2 + CX^2 - 2( OX * CX) cos OXC
OC^2 = 4^2 + 52 - 2 [4 * 2sqrt(13) ] (-2/sqrt (13)
OC^2 = 16 + 52 - 2 [ 4*2 ] (-2)
OC^2 = 16 + 52 + 32
OC^2 = 100
OC = 10
And this is 2/3 of CZ ....so....
(2/3)CZ = 10
CZ = 15