Mathswatch question - Area of a triangle using the sine rule

I have a question missing 1 mark but mathswatch doesnt give it to me
In the screenshot

Working out:
BC = 5a-2b-3a-b = 2a-3b
CD = 3a+b+3a-10b = 6a-9b
CD = 3(BC) or BC = 1÷3(CD)
B,C and D lie on a straight line because CD = 3(BC)
Both BC and CD share point C and are parallel due to CD = 3(BC) so BC and CD lie on a straight line to form BCD

Therefore BC and CD are parallel.They also share a common point A so they lie on the same straight line.
BCD is collinear