A basis for a vector space of dimension n has three properties:
1) The vectors are independent
2) The vectors span the space
3) There are n vectors in the space.
Further, any two of those imply the third.
The simplest way to answer the first question is that a basis for a three dimensional vector space, such as R^3, must contain three vectors, not two.
For the second, rather than using matrices, I would write
a\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}+ b\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}= \begin{bmatrix}a+ b \\ a+ 2b \\ a+ 3b\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
so we must have a+ b= 0, a+ 2b= 0, and a+ 3b= 0. Subtracting the first equation from the second we have b= 0 and then a+ b= a+ 0= 0 so a= 0. The coefficients are a= b= 0 so the vectors are independent.
(Actually,
two vectors are dependent if and only if one is a multiple of the other. Here, it is sufficient to observe that <span style="font-family: 'Verdana'">\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}</span>
is not a multiple of <span style="font-family: 'Verdana'"><span style="font-family: 'Verdana'">\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}</span></span>.)
There are, of course, infinitely many correct answers to
(c) Find a vector, v, such that \{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}, v\} is a basis for R^3. Since those are three vectors it sufficient to find v such that the three vectors are independent or such that they span R^3. You appear to have chosen to find v such that the three vectors are independent but you haven't finished the problem. IF the three vectors were dependent then v would be a linear combination of the other two. In particular, taking the the coefficients to be 1, the sum of the two given vectors is <span style="font-family: 'Verdana'">\{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}+ <div style="text-align: left"><span style="font-family: 'Verdana'"><div style="text-align: left"><span style="font-family: 'Verdana'">\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}= \begin{bmatrix}2 \\ 3 \\ 4\end{bmatrix}</span>​</div><span style="font-family: 'Verdana'"><br />
</span></span>​</div><span style="font-family: 'Verdana'"><span style="font-family: 'Verdana'"><br />
</span></span></span>. To find a vector, v, that is NOT a linear combination, just change one of those components. Change, say, the "4" to "5": v= \begin{bmatrix}2 \\ 3 \\ 5 \end{bmatrix}. The first two components are from "u+ v" while the third is not so v is not any linear combination.
(Since these are 3-vectors, you could also take v to be the "cross product" of the two given vectors. The cross product of two vectors is perpendicular to both so independent of them.)
