The rotation mentioned in this question does not specify clockwise and counterclockwise, so there should be two situations:

(1) When ⊿DCE rotates 60 degrees clockwise, as shown in the figure:

If the extension line of E'H⊥BC is at H, then E 'ch = 60 and CE 'h = 30.

∴ch=( 1/2)ce'=3,e'h=√(e'c^2-ch^2)=3√3；

BE'=√(BH^2+E'H^2)= 14

Let AQ⊥CM be in Q and D 'P ⊥ cm be in P; And CN⊥BE', then ∠CBN=∠ACQ.

And CB = CA∠ CNB =∠ Q = 90, so ⊿ CBN ≌ δ ACQ (AAS), AQ = CN, CQ = BN;

It can also be proved that: ⊿ CPD' ≌ δ E 'NC (AAS), PD' = CN = AQ, CP = E 'n 。

AQ∨PD', then QM/MP=AQ/PD'= 1, so QM=MP.

∴cm=(cp+cq)/2=(e'n+bn)/2=be'/2=7.

According to the area relationship, CB * e 'h = be' * CN, 10 * 3 √ 3 = 14 * CN, CN = 15 √ 3/7.

Therefore: Mn = cm-cn = 7-15 √ 3/7;

(2) Twist ⊿DCE counterclockwise by 60 degrees, and you can also get: cm = 7；;

CN= 15√3/7。 At this time MN=CM+CN=7+ 15√3/7. (Because the examples are similar, I won't go into details. )

So the length of MN is 7- 15√3/7 or 7+ 15√3/7.