The signs are missing from the components of the vectors.
There are a number of ways the signs could have been so 
that the claim would be true.  I will choose one way that 
works.

There are two ways to show that a right triangle is
formed by the vectors, by 
1. showing that the three magnitudes satisfy the Pythagorean 
theorem 
2. showing that the dot product of two of the vectors is zero.
 
I'll use the Pythagorean theorem
method.  Let the three vectors be:

A = 4i-j-3k, B = i-3j-2k, and C = 2i-7k  

If vectors P and Q have a common tail, then the
vector P-Q can have its tip at the tip of P and
its tail at the tip of Q.

Therefore, 

the vector that can have its tip at the
tip of A and its tail at the tip of B is A-B, and
its magnitude is the same as the vector B-A.

A-B = [(4)-(1)]i+[(-1)-(-3)]j+[(-3)-(-2)]k =
[4-1]i+[-1+3]j+[-3+2]k = 3i+2j-k
∥A-B∥ = ∥B-A∥ = 

the vector that can have its tip at the
tip of A and its tail at the tip of C is A-C, and
its magnitude is the same as the vector A-C.

A-C = [(4)-(2)]i+[(-1)-(0)]j+[(-3)-(-7)]k =
[4-2]i+[-1]j+[-3+7]k = 2i-j+4k
∥A-C∥ = ∥C-A∥ = 

the vector that can have its tip at the
tip of B and its tail at the tip of C is B-C, and
its magnitude is the same as the vector B-C.

B-C = [(1)-(2)]i+[(-3)-(0)]j+[(-2)-(-7)]k =
[1-2]i+[-3]j+[-2+7]k = -i-3j+5k
∥B-C∥ = ∥C-B∥ = 

By the inverse of the Pythagorean theorem,

∥A-B∥2+∥A-C∥2 =  ∥B-C∥2

So the 3 vectors joining the tips of A,B, and C form
a right triangle.

Edwin