我遇到一个问题，无法通过parallel包 创建新的图像行
\documentclass[10pt,a4paper]{article}
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\usepackage{amsmath,mathtools}
\usepackage{amsfonts,amssymb}
\usepackage{amsthm,calc}
\usepackage{newtxtext,newtxmath}
\usepackage{caption,cleveref}
\usepackage{graphics}
\usepackage[export]{adjustbox}
\usepackage{lipsum}
\usepackage{parallel}
\usepackage{todonotes}
\usepackage{tcolorbox}
\tcbuselibrary{most,breakable,skins,xparse}
\captionsetup{labelfont=bf, singlelinecheck=true,figureposition=above}% added, had to be here
\captionsetup[table]{singlelinecheck}% added for changing table caption
\newcounter{example}[section]
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\setcounter{equation}{0}}
\newcommand{\unrest}{\setcounter{equation}{\value{foo}}\usetagform{default}}
%---------
\newenvironment{example}[1][]{\parskip=1ex\par\noindent\refstepcounter{example}{\bfseries\theexample}.\enspace#1}{\par\noindent\parskip=1ex}
\newenvironment{solution}[1][]{\textbf{\textit{Solution}}.\enspace\ignorespaces#1\parskip=1ex}{\;$\lhd$\par\noindent\parskip=1ex}
\allowdisplaybreaks
\day 20\month 10 \year 2023
\begin{document}
\lipsum[1-3]
\lipsum[1]
\begin{example}
Evaluate
\[
\iint_R xy \; dA
\]
over the region enclosed\footnote{text} between $y= \frac{1}{2}x$ , $ y= \sqrt{x}$ , $x=2$,$x=4$.
\end{example}
\begin{tcolorbox}[enhanced,breakable,size=title,add to width=2.5cm,right skip=.5cm,left skip=0.2cm,,right=.36cm]
\begin{solution}
\begin{Parallel}[c]{.6\textwidth}{0.34cm}
\ParallelLText{
As we can see from the drawing \cref{fig:dr1} we take a vertical slice and then
% \rule{0pt}{1ex}
\begin{align*}
\int_{2}^{4}\int_{\frac{1}{2} x}^{\sqrt{x}} xy \; dydx
&= \int_{2}^{4} x \frac{y^2}{2}\Big|_{\frac{1}{2} x}^{\sqrt{x}}\\
&=
\frac{1}{2}\int_{2}^{4} {x^2} -{\frac{1}{2} x^3} \; dx\\
&=
\frac{1}{2} \left[\frac{x^3}{3} - \frac{x^4}{16}\right]_{2}^{4}\\
&=\frac{11}{6}
\end{align*}
}
\ParallelRText{
\leavevmode\\%[\baselineskip]
\begin{minipage}[t][0pt][t]{.3\textwidth}
\centering
\includegraphics[valign=t,width=\linewidth]{example-image.pdf}
\captionof{figure}{}
\label{fig:dr1}
\end{minipage}
}
\ParallelPar
\ParallelLText{
As we can see from the drawing \cref{fig:dr2} we take a horizontal slice and then
\begin{align}
\int_{\sqrt{2}}^{2} \int_{y^2}^{2y} xy \tag*{I}\; dx dy\label{eq:ex1}\\
\int_{1}^{\sqrt{2}} \int_{2}^{2y} xy \; dx dy \tag*{II}\label{eq:ex2}
\end{align}
then
\[
\int_{\sqrt{2}}^{2} \int_{y^2}^{2y} xy \; dx dy \;+ \int_{1}^{\sqrt{2}} \int_{2}^{2y} xy \; dx dy \tag*{III} \label{eq:ex3}
\]
\\
in \eqref{eq:ex1}
\begin{align*}
\int_{\sqrt{2}}^{2} \int_{y^2}^{2y} xy \; dx dy
&=\int_{\sqrt{2}}^{2} x(\frac{y^2}{2}) \big|_{y^2}^{2y}\\
&=\frac{1}{2}\int_{\sqrt{2}}^{2} (4y^2-y^4) \\
&=\frac{1}{2}(y^4-\frac{1}{6}y^6)\big|_{\sqrt{2}}^{2}\\
&=\frac{1}{2}\left[ \left(2^4 - \frac{2^6}{6}\right)-\left((\sqrt{2})^4 - \frac{(\sqrt{2})^6}{6}\right) \right] \\
&=\frac{4}{3}\\
\end{align*}
in \eqref{eq:ex2}
\begin{align*}
\int_{1}^{\sqrt{2}} \int_{2}^{2y} xy \; dx dy
&=
\frac{1}{2} \int_{1}^{\sqrt{2}} y(4y^2-2^2)dy \\
&=\frac{(y^2-1)^2}{2}\big|_{1}^{\sqrt{2}} \\
&=\frac{1}{2}
\end{align*}
therefor \eqref{eq:ex3}
\[
\int_{\sqrt{2}}^{2} \int_{y^2}^{2y} xy \; dx dy \;+ \int_{1}^{\sqrt{2}} \int_{2}^{2y} xy \; dx dy \; = \frac{11}{6}
\]
}
\ParallelRText{
\begin{minipage}[t][0pt]{.3\textwidth}
\centering
\includegraphics[valign=t,width=\linewidth]{example-image.pdf}
\captionof{figure}{}
\end{minipage}
\\
\begin{minipage}[t][0pt]{.3\textwidth}
\centering
\includegraphics[valign=t,width=\linewidth]{example-image.pdf}
\captionof{figure}{}
\end{minipage}
}
\ParallelPar
\end{Parallel}
\end{solution}
\end{tcolorbox}
\end{document}