Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)

Well, here's one way to do it at least... 

For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). 

Let P=(4,0) be the projection of B onto the x-axis. 
Let Q=(-3,0) be the projection of C onto the x-axis. 

Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). 

Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). 

Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. 

||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) 
||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) 

Using the Law of Cosines... 
||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) 
||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) 
||a||^2 = 86 - 36 
||a||^2 = 50 
||a|| = sqrt(50) 

Now apply the Law of Sines to find the other two angles. 

||b|| / sin(B) = ||a|| / sin(A) 
sqrt(41) / sin(B) = sqrt(50) / .9080 
(.9080)sqrt(41) / sqrt(50) = sin(B) 
.8222 = sin(B) 
asin(.8222) = B 
55.305 = B 

Two down, one to go... 

||c|| / sin(C) = ||a|| / sin(A) 
sqrt(45) / sin(C) = sqrt(50) / .9080 
(.9080)sqrt(45) / sqrt(50) = sin(C) 
.8614 = sin(C) 
asin(.8614) = C 
59.470 = C 

So your three angles are: 

A = 65.225 
B = 55.305 
C = 59.470