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> Use the definition of continuity to show that $f(x\_1,x\_2,x\_3)= x\_1 -3x\_2 - 2x\_3$ is continuous at each $a \in \mathbb{R}^3$.
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I know the definition says:

$f$ is continuous at $a$ provided that $\forall\epsilon >0$ there exists $\delta >0$ such that $|f(x) - f(a) | < \epsilon$ whenever $\|x-a\|<\delta$.

OR alternatively

$f$ is continuous at $a$ provide that $\forall \epsilon >0$ there exists a $\delta>0$ such that $f(B(a,\delta)) \subset B(f(a),\epsilon)$.

How can I use this to show the given function is continuous at $a$?

**UPDATE:**

Using NotALoner's Hint, I got the following:

Let $\epsilon >0$ be given. To find $\delta >0$ we must find the connection between $|f(x) - f(a)|$ and $||x-a||$.

\begin{align}|f(x) - f(a)| &= | x\_1 -3x\_2 - 2x\_3 - (a\_1 - 3a\_2 - 2a\_3)| \\&= | (x\_1 - a\_1) -3(x\_2 - a\_2) - 2(x\_3 - a\_3)| \\ &\leq ||x-a|| + 3||x-a|| + 2||x-a|| \\ &= 6||x-a||\end{align}

Choose $\delta = \frac{\epsilon}{6}>0$ then for every $x\in \mathbb{R}^3$, whenever $||x-a||< \delta = \frac{\epsilon}{6}$ we have that $|f(x) - f(a)|< 6\frac{\epsilon}{6}=\epsilon$