Show $f:\mathbb{R}^3 \to \mathbb{R}$ is continuous 
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 $a \in \mathbb{R}^3$.

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 $|f(x) - f(a)|< 6\frac{\epsilon}{6}=\epsilon$
 A: *

*Show that $(x_1,x_2,x_3)\mapsto x_1$ is continuous, and likewise $(x_1,x_2,x_3)\mapsto x_2$ and $(x_1,x_2,x_3)\mapsto x_3$.

*Show that a constant function is continuous.

*Show that if $f,g: \mathbb R^3\to \mathbb R$ are both continuous, then $f+g$ and $f\cdot g$ are continuous, too.

*Using (1), (2) and (3), conclude that $-3x_2$ and $x_1-3x_2$ and $-2x_3$ and $x_1-3x_2-2x_3$ are all continuous functions of $x_1$, $x_2$, $x_3$.

*Generalize your method to showing that a polynomial function, even one in several variables, is always continuous.
A: Hint: $$\|f(x_1,x_2,x_3)-f(a_1,a_2,a_3)\| = \left|(x_1-a_1)-3(x_2-a_2)-2(x_3-a_3)\right| \leq |x_1-a_1|+3|x_2-a_2|+2|x_3-a_3| \leq 6\sqrt{(x_1-a_1)^2+(x_2-a_2)^2+(x_3-a_3)^2}=6\|(x_1,x_2,x_3)-(a_1,a_2,a_3)\|$$
A: Consider $x=(x_1,x_2,x_3)$ and $a=(a_1,a_2,a_3)$. Then
\begin{align}
|f(x)-f(a)|
&=|f(x_1,x_2,x_3)-f(a_1,a_2,a_3)|\\
&=|(x_1-a_1)-2(x_2-a_2)-3(x_3-a_3)|\\
&\le|x_1-a_1|+2|x_2-a_2|+3|x_3-a_3|
\end{align}
If $\|x-a\|<\frac{\varepsilon}{6}$ you certainly have
$$
|x_i-a_i|<\frac{\varepsilon}{6}\qquad(i=1,2,3)
$$
so
$$
|f(x)-f(a)|<\frac{\varepsilon}{6}+2\frac{\varepsilon}{6}+3\frac{\varepsilon}{6}=\varepsilon
$$
Note that the same argument shows that any linear function $f\colon\mathbb{R}^3\to\mathbb{R}$ is continuous, because it has the form
$$
f(x_1,x_2,x_3)=Ax_1+Bx_2+Cx_3
$$
and it's sufficient to take
$$
\delta=\frac{\varepsilon}{|A|+|B|+|C|}
$$
(provided the function $f$ is not the constant zero, when there's of course nothing to prove).
