If $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ are unit vectors, then find the largest possible value of
\[\|\mathbf{a} - \mathbf{b}\|^2 + \|\mathbf{a} - \mathbf{c}\|^2 + \|\mathbf{b} - \mathbf{c}\|^2.\]Note: A unit vector is a vector of magnitude 1.
We can write
\begin{align*}
\|\mathbf{a} - \mathbf{b}\|^2 &= (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \\
&= \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \\
&= \|\mathbf{a}\|^2 - 2 \mathbf{a} \cdot \mathbf{b} + \|\mathbf{b}\|^2 \\
&= 2 - 2 \mathbf{a} \cdot \mathbf{b}.
\end{align*}Similarly, $\|\mathbf{a} - \mathbf{c}\|^2 = 2 - 2 \mathbf{a} \cdot \mathbf{c}$ and $\|\mathbf{b} - \mathbf{c}\|^2 = 2 - 2 \mathbf{b} \cdot \mathbf{c},$ so
\[\|\mathbf{a} - \mathbf{b}\|^2 + \|\mathbf{a} - \mathbf{c}\|^2 + \|\mathbf{b} - \mathbf{c}\|^2 = 6 - 2 (\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}).\]Now,
\[\|\mathbf{a} + \mathbf{b} + \mathbf{c}\|^2 \ge 0.\]We can expand this as
\[\|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 + \|\mathbf{c}\|^2 + 2 \mathbf{a} \cdot \mathbf{b} + 2 \mathbf{a} \cdot \mathbf{c} + 2 \mathbf{b} \cdot \mathbf{c} \ge 0.\]Then $2 (\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}) \ge -3,$ so
\[\|\mathbf{a} - \mathbf{b}\|^2 + \|\mathbf{a} - \mathbf{c}\|^2 + \|\mathbf{b} - \mathbf{c}\|^2 = 6 - 2 (\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}) \le 9.\]Equality occurs when $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ are equally spaced on a circle with radius 1 (where $\|\mathbf{a} - \mathbf{b}\| = \|\mathbf{a} - \mathbf{c}\| = \|\mathbf{b} - \mathbf{c}\| = \sqrt{3}$), so the largest possible value is $\boxed{9}.$

[asy]
unitsize(2 cm);

pair A, B, C;

A = dir(20);
B = dir(20 + 120);
C = dir(20 + 240);

//draw((-1.5,0)--(1.5,0));
//draw((0,-1.5)--(0,1.5));
draw(Circle((0,0),1));
draw((0,0)--A,Arrow(6));
draw((0,0)--B,Arrow(6));
draw((0,0)--C,Arrow(6));
draw(A--B--C--cycle,dashed);

label("$\mathbf{a}$", A, A);
label("$\mathbf{b}$", B, B);
label("$\mathbf{c}$", C, C);
[/asy]
We can write
\begin{align*}
\|\mathbf{a} - \mathbf{b}\|^2 &= (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \\
&= \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \\
&= \|\mathbf{a}\|^2 - 2 \mathbf{a} \cdot \mathbf{b} + \|\mathbf{b}\|^2 \\
&= 2 - 2 \mathbf{a} \cdot \mathbf{b}.
\end{align*}Similarly, $\|\mathbf{a} - \mathbf{c}\|^2 = 2 - 2 \mathbf{a} \cdot \mathbf{c}$ and $\|\mathbf{b} - \mathbf{c}\|^2 = 2 - 2 \mathbf{b} \cdot \mathbf{c},$ so
\[\|\mathbf{a} - \mathbf{b}\|^2 + \|\mathbf{a} - \mathbf{c}\|^2 + \|\mathbf{b} - \mathbf{c}\|^2 = 6 - 2 (\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}).\]Now,
\[\|\mathbf{a} + \mathbf{b} + \mathbf{c}\|^2 \ge 0.\]We can expand this as
\[\|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 + \|\mathbf{c}\|^2 + 2 \mathbf{a} \cdot \mathbf{b} + 2 \mathbf{a} \cdot \mathbf{c} + 2 \mathbf{b} \cdot \mathbf{c} \ge 0.\]Then $2 (\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}) \ge -3,$ so
\[\|\mathbf{a} - \mathbf{b}\|^2 + \|\mathbf{a} - \mathbf{c}\|^2 + \|\mathbf{b} - \mathbf{c}\|^2 = 6 - 2 (\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}) \le 9.\]Equality occurs when $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ are equally spaced on a circle with radius 1 (where $\|\mathbf{a} - \mathbf{b}\| = \|\mathbf{a} - \mathbf{c}\| = \|\mathbf{b} - \mathbf{c}\| = \sqrt{3}$), so the largest possible value is $\boxed{9}.$

[asy]
unitsize(2 cm);

pair A, B, C;

A = dir(20);
B = dir(20 + 120);
C = dir(20 + 240);

//draw((-1.5,0)--(1.5,0));
//draw((0,-1.5)--(0,1.5));
draw(Circle((0,0),1));
draw((0,0)--A,Arrow(6));
draw((0,0)--B,Arrow(6));
draw((0,0)--C,Arrow(6));
draw(A--B--C--cycle,dashed);

label("$\mathbf{a}$", A, A);
label("$\mathbf{b}$", B, B);
label("$\mathbf{c}$", C, C);
[/asy]
The final answer is $$\boxed{9}$$. I hope it is correct.