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 :

$\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]$

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 :

$\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]$

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 :

$\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]$

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

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