Answer:
a. Line
b. Plane
c. All of R^3
Step-by-step explanation:
In order to answer this question, we need to study the linear independence between the vectors : 
1 - A set of three linearly independent vectors in R^3 generates R^3.
2 - A set of two linearly independent vectors in R^3 generates a plane.
3 - A set of one vector in R^3 generates a line. 
The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :
a. Let's put the vectors into the rows of a matrix : 
![\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%265%26-3%5C%5C6%26-15%269%5C%5C-10%2625%26-15%5Cend%7Barray%7D%5Cright%5D) ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒
 ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒
 ![\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%265%26-3%5C%5C0%260%260%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).
We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).
At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.
b. Again, let's put the vectors into the rows of a matrix : 
![\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%260%5C%5C1%261%261%5C%5C4%265%263%5Cend%7Barray%7D%5Cright%5D) ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒
 ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒
![\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C0%261%26-1%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector). 
c. Finally :
![\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%263%5C%5C0%261%262%5C%5C1%261%260%5Cend%7Barray%7D%5Cright%5D) ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒
 ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒
![\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%261%262%5C%5C0%260%263%5Cend%7Barray%7D%5Cright%5D)
The set is linearly independent so the set of all linear combination of the set c. is all of R^3.