I get the same answer as your prof, (by a different method from the one that you have used).
I think that your value for |AxB| is wrong.
alright so I am having doubt in my professors answer to this question or maybe I am doing something wrong so before i go speak with him, i wanna ask some other professionals
Find the distance of the point P = (1,2,-3) to the line: (1,-2,1) + t(2,3,4)
The vector from (1,2,-3) to (1,-2,1) is (0,4, -4) = A
and the direction vector is (2,3,4) = B
so sinθ = |AxB|/ [|A|][|B|]
where |AxB| = 12
|A| = √32
|B| = √29
so |A|sinθ = 12/√29
however my profs answer is √(32 - 16/29)
sorry i din't latex it i was in a rush
The prof's method was probably to minimise the distance from the point (1+2t,–2+3t,1+4t) to the point (1,2,–3) by completing the square. That distance is , which also comes out as when t has the minimising value –4/29.
Personally, I prefer the cross product method to the minimisation method (provided that it's done correctly).