Given,

AC is perpendicular to CE

∠A: ∠B: ∠C = 3: 2: 1

Let,

∠A = 3k

∠B = 2k

∠C = k

∠A + ∠B + ∠C = 180^o

3k + 2k + k = 180^o

6k = 180^o

k = 30^o

Therefore,

∠A = 3k = 90^o

∠B = 2k = 60^o

∠C = k = 30^o

Now,

∠C + ∠ACE + ∠ECD = 180^o (Linear pair)

30^o + 90⁰ + ∠ECD = 180^o

∠ECD = 180^o – 120^o

= 60^o