Princess Eugenie’s father in law is in a “serious but stable” condition in hospital after testing positive for coronavirus.
- Jack Brooksbank’s 71-year-old father, George, is currently in intensive care being treated for the virus.
- Eugenie’s mother-in-law, Nicola, 66, has also contracted the virus but is recovering at home.
- This folloes royal news that Prince Charles, had tested positive for covid-19 – as the world continues to be gripped by the pandemic.
Despite his diagnosis, Charles, 71 – who had mild symptoms – self-isolated for seven days and is now on the road to recovery.
While Princess Eugenie is yet to mention her in-laws’ experience of coronavirus on social media, she and her mother, the Duchess of York, have been heavily involved in supporting NHS workers during these unprecedented times by delivering care packages to various hospitals in London.
And an insider said the mother-daughter duo “wanted to do anything they could to help” amid the COVID-19 crisis.
The care packages include essentials such as hand wash, anti-bacterial wipes and dried pasta as well as sweet treats and confectionary – thanks to The Duchess’ good friend and confectionary entrepreneur, Hermann Bühlbecker.
Revealing how he assisted in their mission, Hermann said, “A few days ago Fergie, Duchess of York, and her daughter received a large delivery of Lambertz pastries (over 100 kg) at the Royal Lodge.
“This is intended for the doctors and staff of London hospitals, who are currently working up to their limits. The Duchess called us about this; she sent us the pictures when the shipment arrived.”
Though it appears Eugenie and her mother, Sarah Ferguson, are self-isolating together, Princess Beatrice seems to have taken the decision not to return home to isolate.
While it is unclear where Princess Beatrice and her fiancé, Edoardo Mapelli Mozzi, are currently residing, the couple have taken the “very sad” decision to postpone their wedding – which was due to take place in May – over the coronavirus outbreak.
From our sister site, GoodtoKnow.