$re = '~<a\s[^>]*?href\s*=\s*\\\\\'https?://site\.ru/[^>]+>\s*<img.+?</a>~s';
$str = '<a class=\\\'fancybox\\\' href=\\\'http://site.ru/upload/manuals/3V/image_2022_01_28T12_03_17_757Z.jpg\\\' rel=\\\'group\\\'><img src=\\\'http://site.ru/upload/manuals/3V/image_2022_01_28T12_03_17_757Z.jpg\\\' width=\\\'400px\\\'></img></a>
<a class=\\\'fancybox\\\' href=\\\'http://site2.ru/upload/manuals/3V/image_2022_01_28T12_03_17_757Z.jpg\\\' rel=\\\'group\\\'><img src=\\\'http://site2.ru/upload/manuals/3V/image_2022_01_28T12_03_17_757Z.jpg\\\' width=\\\'400px\\\'></img></a>
<a class=\\\'fancybox\\\' href=\\\'http://site.ru/upload/manuals/3V/image_2022_01_28T12_03_17_757Z.jpg\\\' rel=\\\'group\\\'><img src=\\\'http://site.ru/upload/manuals/3V/image_2022_01_28T12_03_17_757Z.jpg\\\' width=\\\'400px\\\'></img></a>';
$subst = "";
$result = preg_replace($re, $subst, $str);
echo "The result of the substitution is ".$result;
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for PHP, please visit: http://php.net/manual/en/ref.pcre.php