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Please, only hints and suggestions

I've tried many things with all fails.

I knew that this set of vectors $\{v_1, v_2, v_3 \}$ is Linearly Independent. So I have to find some vector $v$ such that

$c_1v_1 + c_2v_2 + c_3v_3 + c_4v = 0$ but one of them isn't 0.

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    Why do you think that v1,v2,v3 is linear dependent? – 2017-01-15
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    Can you solve the equation for $v$? – 2017-01-15
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    @benji, like $v = \frac{c_1v_1 + c_2v_2 + c_3v_3}{c_4}$? – 2017-01-16

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Let $v=[a\;b\;c\;d]^T$ and consider Gaussian elimination on the matrix \begin{align} \begin{bmatrix} 1 & 0 & 1 & a\\ -1 & 1 & -2 & b\\ 0 & 1 & -2 & c\\ 2 & 0 & 2 & d \end{bmatrix} &\to \begin{bmatrix} 1 & 0 & 1 & a\\ 0 & 1 & -1 & b+a\\ 0 & 1 & -2 & c\\ 0 & 0 & 0 & d-2a \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 & 1 & a\\ 0 & 1 & -1 & b+a\\ 0 & 0 & -1 & c-b-a\\ 0 & 0 & 0 & d-2a \end{bmatrix} \end{align} Thus you see that the vector should satisfy $d=2a$, which is the condition for the linear system $$ c_1v_1+c_2v_2+c_3v_3+c_4v=0 $$ to have a non trivial solution.

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    very fast. keep it going. – 2017-01-16
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    I am having some trouble. We solve $c_1v_2 + c_2v_2 + c_3v_3 + c_4v_4$? Using Augmented matrices? – 2017-01-17
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    Also, you didnt complete the RREF? Is it valid still? – 2017-01-17
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    @Amad27 At that stage, we already know what the pivot columns are, the RREF wold just tell us how to write the vector $v$ as a linear combination of the three vectors (provided $ d=2a$). We're essentially solving $v=c_1v_1+c_2v_2+c_3v_3$. – 2017-01-17
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    @egreg, also is this matrix augmented? Is the augmented second part $[a, b, c, d]$? – 2017-01-23
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    @Amad27 You can see it as an augmented matrix, if you deem so. – 2017-01-23
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    @egreg, is this another way to see it: You are solving such that $v$ is a linear combination of $v1, v_2, v_3$. So you are solving $c_1 v_1 + c_2v_2 + c_3v_3 = v$? – 2017-01-23
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    @Amad27 Yes, that can be an interpretation as good as “find the dimension of the span of $\{v_1,v_2,v_3,v\}$. – 2017-01-23