Find the requirement for a list of vectors to be linearly independent 
Suppose there is a list $((r,1,1),(0,s,1),(1,1,t))$(r,s,t$\in \mathbb Q$) and then find the requirement needed this list to be linearly independent in $\mathbb R^3$.

I find that this question is not very approachable to me. I know the requirement for a list being linearly independent should be:  $a(r,1,1)+b(0,s,1)+c(1,1,t)=0 \iff a=b=c=0$
Then
$\begin{array}
    ar+c=0  \\
    a+bs+c=0 \\                
    a+b+ct=0
\end{array}$
Then the aim becomes if these three equations are satisfied, then it must be a=b=c=0. Then I think all I need to do is to find the proper value of r,s, and t such that they will force a=b=c=0.
However,(if my idea is correct) I don't know how to find such r,s, and t. The way I tried is to use the equivalent negation, i.e. if $a\neq0$ or $b\neq0$ or $c\neq0$. Then one of these equations can't be zero if r,s,t are rational numbers, but I failed. Any suggestions?

You have the right idea looking for negation, just have some details off - we have the associated set of equations associated with the scaled sums of the vectors equaling zero given by $$ \begin{matrix}ar + c = 0 \\ a + bs + c = 0 \\ a + b + ct = 0. \end{matrix}$$ From here, notice that $$ ar + c = 0 \Rightarrow c = -ar$$ which tells us something important - if $a = 0$, then $c = 0$ - however, note that this would imply that $b$ would have to be zero and as such the vectors would be linearly independent! As such, continuing from here we can assume that $a \not= 0$. Now, let's handle the cases where either $r = 0$ or $r \not= 0$ separately :
First, the case where $r = 0$. Then $c = 0$, so we get from the third equation $$ a + b = 0 \Rightarrow a = -b $$ and plugging that into the second equation gives us $$ a + bs = 0 \Rightarrow -b + bs = 0 \Rightarrow b(s - 1) = 0 $$ which means that if $b \not= 0$, $s = 1$. However, if $b = 0$ then $a = -b = 0$, and hence if $r = 0$ we have that the set of vectors given is linearly independant if and only if $s = 1$.
Now, we handle the case where $r \not= 0$. In this case, note that we have $$c = -ar \Rightarrow a = -c(\frac{1}{r})$$ which we weren't able to derive before, since $\frac{1}{r}$ is undefined for $r = 0$. Now, plugging this into the third equation gives us $$ a + b + ct = 0 \Rightarrow -c(\frac{1}{r}) + b + ct = 0 \Rightarrow b + c(t - \frac{1}{r}) = 0 \Rightarrow b = c(\frac{1}{r} - t) $$ and a substitution of both the identity for $a$ and $b$ into the second equation gives $$ a + bs + c = 0 \Rightarrow c(-\frac{1}{r}) + c(\frac{s}{r} - st) + c = 0 \Rightarrow c(1 + \frac{s - 1}{r} - st) = 0 \Rightarrow c(r + s - 1 - str) = 0$$ with the final implication being particularly useful in this case since $r \not= 0$. Now, since we know that if $c = 0$ then $a = 0$ (because $r \not= 0$) then this is equivalent to the statement that $$ r + s - 1 - str = 0. $$ From here, notice something interesting - if $r = 0$ then $$r + s - 1 - str = 0 \Rightarrow s - 1 = 0 \Rightarrow s = 1$$ which means that we can summarize the entire discussion up to this point (including the case where $r = 0$) by the following: the vectors $\lbrace (r, 1, 1), (0, s, 1), (1, 1, t) \rbrace$ are linearly independant if and only if $r + s - 1 - str \not= 0$.
While this post (and problem) was very computation heavy, this problem comes with a lesson for other problems like it - from a setup, you will be able to extract givens from other assumptions (in this case, is was that $c = -ar$ under the assumption that the sums of the scaled vectors equal $0$). From there givens, think extremely deeply about what such a given would imply (in this case it was that if $a = 0$ then $a = b = c = 0$), and see how those findings interact with other assumptions.
