New Delhi [India], Mar 3 : After a passenger who travelled by the February 25 Vienna-Delhi flight tested positive for COVID-19, Air India has asked the remaining passengers on the flight to follow the protocol notified by the Union Ministry of Health and Family Welfare for prevention of coronavirus spread.
"This is for the attention of passengers who flew on AI154 Vienna-Delhi of 25th Feb' 20. One of the passengers has tested positive for coronavirus. Please follow the protocol notified by the Ministry of Health regarding coronavirus," Air India stated.
All crew members of the flight have been asked to self-quarantine for a period of 14 days, an Air India official told .
"All crew members of the February 25 Vienna-Delhi flight have been asked to stay in isolation at their respective homes for 14 days. A male passenger, who travelled from Vienna to Delhi by this flight, was found to be infected with coronavirus," said the official on Monday.
The Union Ministry of Health and Family Welfare has issued detailed guidelines regarding the prevention of the infection and what to do if a person shows signs of COVID-19.
The Ministry of Health and Family Welfare earlier announced that two cases of COVID-19 have been detected in the country including the one in the national capital with a travel history from Italy. The other case has been reported from Telangana. Both the patients are stable and are being closely monitored.
Earlier, three cases of COVID-19 were reported in Kerala. All three have now been discharged following treatment.
( With inputs from ANI )