Look at the diagram.

ID, IE, IF are radii of incircle and perpendiculars on to the sides from incenter I.

Triangles AIE, AIF are similar as AI is common, IE=IF, and angles IEA = IFA = 90 deg. Also, AE=AF as tangents from same point A to incircle.

Similarly, triangles BIF, BID are similar. Also, CID, and CIE are similar.

Hence, angle AIF=AIE= z.

In the quadrilateral AFIE, angle A = 360-(90+90+FIE) = 180 - 2 z

Similarly from quadrilateral BFID, we get angle B = 180 - 2 x

From quadrilateral CDIE, we get angle C = 180 - 2 y

We are given that BIC : CIA : AIB = 5 : 6 : 7

x+y : y+z : z+x = 5 : 6 : 7

(x+y)6 = (y+z)5 => y + 6x = 5z

(x+y)7 = (z+x) 5 => 2x + 7y = 5z

solving these simultaneous equations, we get 6 y = 4 x, or 3 y = 2 x

and 10 y = 5z, z = 2y

angle A = 180 - 2 z = 180 - 4y, angle B = 180 - 2 x = 180 - 3y

angle C = 180 - 2 y

Their sum = A+B+C = 180 => 9 y = 360, y = 40 deg.

angle A = 20 deg, angle B = 60 deg angle C = 100 deg.

A : B : C = 1 : 3 : 5