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Is the set of vectors linearly dependent? If so write one of them as a combination of others.

Answer: By the zero vector theorem, since the set contains the zero vectors, it must be that the set is linearly dependent.

Find one vector as a combination of others.

I am unable to do that, but it should work?

I am still getting all coefficients to be $0$?

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    It's still a linear combination if the coefficients are zero. – 2017-01-09
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    @juanarroyo, but that would make it linearly independent – 2017-01-09
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    The relation you have is $0\times \vec {v_1}+0\times \vec {v_2}=1\times \vec {v_3}$. Sure, two of the coefficients are $0$ but not all three. – 2017-01-09
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    Vectors ${v_i}$ are linearly independent if $\sum c_i v_i=0 \implies$ all the coefficients are zero. That condition is false for these vectors. – 2017-01-09

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$0\cdot \begin{pmatrix} 1\\ -3\\ -2 \end{pmatrix} +0 \cdot \begin{pmatrix} 4\\ 6\\ 1 \end{pmatrix}+k \cdot \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

The equation holds for any $k \ne 0$. Then the set of vectors are linearly dependent.

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    @Amad27: is it clear? – 2017-01-21
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Any set of vectors containing the zero vector is linearly dependent.

The zero vector, $0$, can be shown to be a linear combination of any other vectors with coefficients all 0, except $a$ can be an arbitrary number and this criteria is met, so the zero vector being in a set of vectors causes them to become a linearly dependent set. There then exists a dependence relation.

Recall, a finite set {$x_i $} of vectors is linearly dependent if there exists a corresponding set {$a_i $} of scalars, not all zero, such that $\sum_i a_ix_i = 0$.