vector space R³ 
Determining the three vectors are linearly dependent or independent
  $$u = (1, 2, -3), v = (3, 1, -2), w =(5, -5, 6). $$

Is looking through determinant enough to check whether linearly dependent or independent?

I know that determinant of the matrix is zero. So, Can I say it is linearly independent because of that determinant is zero ? To find the relation between $u$, $v$, and $w$ we look for constants a, b, and c. When I apply the echelon form I get  $\begin{bmatrix}1 & 0 & -4\\0& 1 & 3\\0 &0 &0\end{bmatrix}$. Up to here I understand, but I'm stuck here

how can we select $c = 1$ ? How come $b = -3c$ and $a = -3b - 5c $ ? Written $u$, $v$, and $w$ are linearly independent. But, the determinant is zero? I'm confused.

You have some special vectors $u$, $v$, and $w$ that their linear dependence or independence is of interest. Let us put each of them into the rows of a matrix so that we have
$${\left[ {\begin{array}{*{20}{c}}
u\\
v\\
w
\end{array}} \right]_{3 \times 3}} = \left[ {\begin{array}{*{20}{c}}
1&2&{ - 3}\\
3&1&{ - 2}\\
5&{ - 5}&6
\end{array}} \right]$$
Now do the elementary row operations (ERO) to get the echelon form:
$$\left[ {\begin{array}{*{20}{c}}
1&2&{ - 3}\\
0&{ - 5}&7\\
0&0&0
\end{array}} \right]$$
Now can you tell me that what a zero row means? We did ERO which means we just computed some linear combinations of the vectors and we ended up we zero vector! Yes! This means that your vectors are linearly dependent.
