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By solving vectors $(4, -2, -2)$, $(3, -2, -3)$ and $(-5, 4, 7)$, I get:

$x_1 = -x_3$ ,

$x_2 = 3x_3$ ,

$x_3$ = free

How I can find that these vectors span a line in $ \mathbb{R}^3$ or a plane in $ \mathbb{R}^3$ or $\mathbb{R}^3$ itself.

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    The number of free variables determine the sort of object. Hint: how many directions do you need for a line? – 2017-02-06
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    There is 1 free variable, so the *solution space* is a line in $R^3$. The vectors span a $(3-1)=2$-dimensional subspace in $R^3$, which is a plane. – 2017-02-06
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    @woogie, if the three vectors are linearly independent, shouldnt it span $R^3$? – 2017-02-06
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    Yes, 3 linearly independent vectors in $R^3$ will span $R^3$. – 2017-02-06

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Let $$v_1=(4,-2,-2)$$ $$v_2=(3,-2,-3)$$ $$v_3=(-5,4,7)$$ Note that $v_1-3v_2=v_3$. We have $2$ linearly independent vectors $v_1$ and $v_2$, and $1$ dependent vector $v_3$.
$2$ linearly independent vectors in $\mathbb{R}^3$ span a plane.

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    So, these vectors span a plane in R3. Right ? – 2017-02-06
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    Yes, they do. $ $ – 2017-02-06
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    Thank You for your help :) – 2017-02-06
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    No problem. You can choose to accept your favorite answers by toggling a check mark next to it. – 2017-02-06
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To span all of $\mathbb{R}^{3}$, we need $3$ linearly independent vectors (non-parallel, non-coplanar).

You just said those vectors depend on eachother.

$x_1$ can be written in terms of $x_3$, and $x_2$ can also be written in terms of $x_3$. But notice that $x^3$ is only dependent on itself, and as such, it is the only vector that matters. If you only have one vector, you span a line.