Answer
$\sqrt{6}$
Work Step by Step
I know that for the vector $v=\begin{bmatrix}
v_{1} \\
v_{2} \\
\vdots\\
v_{n}
\end{bmatrix}$
$||v||=\sqrt{v_1^2+v_2^2+...+v_n^2}$
The distance of two vectors $u$ and $v$ is: $d=||u-v||$.
Hence: $d=||\begin{bmatrix}
1 \\
2 \\
3 \\
\end{bmatrix}-\begin{bmatrix}
2 \\
3 \\
1\\
\end{bmatrix}||=||\begin{bmatrix}
-1 \\
-1 \\
2\\
\end{bmatrix}||=\sqrt{(-1)^2+(-1)^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$
