Ifthe three given vectors are independent, then theyarea basis for their own span and the dimension is 3. If they are not, then one of the vectors can be written as a linear combination of the other two and so those two form a basis and the dimension is 2. (It clear that these vectors are not all multiples of one another so the dimension is not 1.)

So start by checking if they are independent or not: do there exist numbers, a, b, and c such that

That is the same as

and that is the same as the three equations a+ b- c= 0, 2a+ b= 0, and 3a- b+ 5c= 0. Obviously a= b= c= 0 is a solution. If it is theonlysolution, then the three vectors form a basis and the space is three dimensional.

If there exist another solution, we can solve for one of the vectors in terms of the other two. If we had with, at least, c non-zero, then we can write so . Since , , and span the space every vector in it can be written as a linear combination, . With we can replace and get . Then is a basis and the space has dimension 2.

That's the idea. I'll leave the actual calculation to you.