Three. (20 points) Let the edge length of the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ be 1. Find a point $P$ on the diagonal $B D_{1}$ and a point $Q$ on the base quadrilateral $A B C D$ such that the value of $P Q+P C_{1}$ is maximized.

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$$
\begin{array}{l}
\text { Three, as shown in Figure 6, let } \\
P B=x, \angle D_{1} B C_{1} \\
=\alpha \text {. Then } \\
\cos \alpha=\frac{\sqrt{2}}{\sqrt{3}} \\
=\frac{\sqrt{6}}{3} \text {, } \\
P C_{1}^{2} \\
=x^{2}+2-2 \sqrt{2} x \cos \alpha \\
=x^{2}+2-\frac{4 \sqrt{3}}{3} x \text {. } \\
\end{array}
$$

Let $A C$ intersect $B D$ at point $O$, and draw $P M \perp$ plane $A B C D$ at $M$. Then
$$
\frac{P M}{1}=\frac{B M}{\sqrt{2}}=\frac{x}{\sqrt{3}} .
$$

Therefore, $P M=\frac{\sqrt{3}}{3} x, B M=\frac{\sqrt{6}}{3} x$.
When $x \leqslant \frac{\sqrt{3}}{2}$, $M$ is on segment $O B$, at this time,
$\angle M A D>45^{\circ}=\angle A D M$.
Thus, $M C=M A\frac{\sqrt{3}}{2}$, when $M$ is on segment $O D$. By symmetry, the circle with $M$ as the center and $M B$ as the radius covers the square $A B C D$. Therefore, it covers point $Q$. Thus, $M Q \leqslant M B$. Therefore, $P Q \leqslant P B$.
Hence $P Q^{2} \leqslant P B^{2}=P M^{2}+B M^{2}$
$$
=\frac{1}{3} x^{2}+\left(\frac{\sqrt{6}}{3} x\right)^{2}=x^{2} \text {. }
$$

Then $\left(P Q+P C_{1}\right)^{2} \leqslant 2\left(P Q^{2}+P C_{1}^{2}\right)$
$\leqslant 2 x^{2}+2 x^{2}+4-\frac{8 \sqrt{3}}{3} x=4 x^{2}-\frac{8 \sqrt{3}}{3} x+4$.
Since $\frac{\sqrt{3}}{2}<x \leqslant \sqrt{3}$, the function $4 x^{2}-\frac{8 \sqrt{3}}{3} x+4$ is increasing on $\left(\frac{\sqrt{3}}{2}, \sqrt{3}\right]$, so, $4 x^{2}-\frac{8 \sqrt{3}}{3} x+4 \leqslant 8$, i.e.,
$$
\left(P Q+P C_{1}\right)^{2} \leqslant 8 \Rightarrow P Q+P C_{1} \leqslant 2 \sqrt{2} .
$$

When and only when $x=\sqrt{3}$ and $P Q=P C_{1}$, the equality holds.
But these two conditions do not hold simultaneously.
Therefore, $P Q+P C_{1}<2 \sqrt{2}$.
In summary, when $P$ is at point $B$ and $Q$ is at point $D$, the value of $P Q+P C_{1}$ is maximized, with the maximum value being $2 \sqrt{2}$.