11 (22 points) Find all solutions to the equation $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x$, where $[a]$ denotes the greatest integer less than or equal to $a$.

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(11 Method 1: From the equation, we know that the solution $x$ is an integer. Let $x=42 p+q$, where $p \in \mathbf{Z}$, $q \in\{0,1, \cdots, 41\}$. Then
$$
\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x \Leftrightarrow p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q .
$$

Therefore, the solution set of the equation is
$$
\begin{aligned}
& \left\{\left.42\left(\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q\right)+q \right\rvert\, q=0,1, \cdots, 41\right\} \\
= & \{0,-6,-12,-14,-18,-20,-21,-24,-26,-27, \\
& -28,-30,-32,-33,-34,-35,-36,-38,-39, \\
& -40,-41,-44,-45,-46,-47,-49,-50,-51, \\
& -52,-53,-55,-57,-58,-59,-61,-64,-65, \\
& -67,-71,-73,-79,-85\} .
\end{aligned}
$$

Method 2: From the equation, we know that the solution $x$ is an integer. By $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}=\frac{41}{42}<1$ we know $x \leqslant$ set
$$
a=-x(\bmod 2), b=-x(\bmod 3), c=-x(\bmod 7),
$$

then there exists $d \in \mathbf{Z}$ such that $-x=21 a+14 b+6 c+42 d$. Therefore,
$$
\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x \Leftrightarrow d=0 .
$$

The solution set of the equation is
$$
\begin{aligned}
& \{0,-21\} \oplus\{0,-14,-28\} \\
& \oplus\{0,-6,-12,-18,-24,-30,-36\} \\
= & \{0,-6,-12,-14,-18,-20,-21,-24,-26,-27, \\
& -28,-30,-32,-33,-34,-35,-36,-38,-39, \\
& -40,-41,-44,-45,-46,-47,-49,-50,-51, \\
& -52,-53,-55,-57,-58,-59,-61,-64,-65, \\
& -67,-71,-73,-79,-85\} .
\end{aligned}
$$