Imagine all the points

Suppose the center of sphere is $O,$ and the midpoint of $AC$ is $E.$ Consider the triangle $EKS.$ It is a right triangle, with $\angle KES=60^\circ.$ The point $O$ lies on $KS,$ at the distance $1$ from $K$ and from the side $SE.$ Therefore, $EO$ is the angle bisector of $\angle KES,$ thus $\angle KEO=30^\circ.$ This implies that $|KE|=\sqrt{3},$ thus $|SE|=2\sqrt{3}$ and $|KS|=3.$

In the equilateral triangle $ABC,$ the segment $KE$ is the in-radius. Because $\angle AKE = 60^\circ,$ $|AE|=\sqrt{3} \cdot \tan 60^\circ = 3.$ So $|AC|=6.$

Because a plane through $M$ and $N$ is tangent to the sphere at a point equidistant from $M$ and $N,$ we can conclude (e.g., using the Pythagorean Theorem) that $|OM|=|ON|.$ Therefore, $|KM|=|KN|.$ Disregard for a moment the condition $|MN|=5.$ As a point $M$ moves from $A$ to $B,$ its distance from $K$ first decreases from $2\sqrt{3}$ to $\sqrt{3},$ then increases back to $2\sqrt{3},$ after it passes the midpoint of $AB.$ For every such $M$ there are two points $N$ on $BC$ at the same distance from $K$ (they coincide when $M$ is the midpoint of $AB$ ). One of them is symmetric to $M$ about the line $BK$ . The other one has $\angle MKN=120^\circ.$

We will show that the second possibility is impossible, given the other conditions of the problem. Indeed, if $|MN|=5$ and $\angle MKN =120^\circ,$ the distance from the midpoint of $MN$ to $K$ is $\frac{5}{2\sqrt{3}}.$ Because this is greater than $1,$ the plane through $MN$ that is tangent to the sphere (and is different from the plane $ABC$ ) will intersect $OK$ beyond $O,$ rather than beyond $K.$

So, $M$ and $N$ are symmetric about the line $BK.$ Note that the midpoint of $MN$ lies in the plane $SKE.$ Let us call it $F.$ Because $|BK|:|KE|=2:1,$ the length of the cross-section of $ABC$ passing through $K$ parallel to $AC$ equals $\frac{2}{3}\cdot 6 =4.$ Because $|MN|=5=\frac{4+6}{2},$ $F$ is the midpoint of $KE.$

Now we do some calculations in the plane $SKE$ . Suppose $\angle KOF=\alpha.$ Suppose $G$ is the point of the plane through $MN,$ where it is tangent to the sphere. Then $\angle FOG=\angle KOF= \alpha,$ so $\angle KFD=180^\circ-2(90^\circ-\alpha))=2\alpha.$ Since $\tan \alpha = \frac{\sqrt{3}}{2},$ $\tan (2\alpha )=\frac{\sqrt{3}}{1-\frac{3}{4}}=4\sqrt{3}.$ So $|DK|=4\sqrt{3}\cdot \frac{\sqrt{3}}{2} = 6.$

To finish up, $|CK|=2\sqrt{3}.$ By the Pythagorean theorem for the triangle $DKC$ , $|CD|^2=36+12=48.$