New Delhi, June 1 (ANI): South African President Dr. Jacob Zuma will undertake a three-day visit of India from Wednesday (June 2).
Dr.Zuma will first visit Mumbai for a day on Wednesday before heading for the Indian capital on Thursday. He will arrive in New Delhi at 5 p.m.
On Friday, Dr. Zuma will be given a ceremonial reception in the forecourt of Rashtrapati Bhawan. He will be received by President Pratibha Devisingh Patil and Prime Minister Dr. Manmohan Singh.
At 9.30 a.m., he will proceed to lay a wreath at the memorial of ahatma Gandhi, Father of The Nation at Rajghat
Minister of State for External Affairs Preneet Kaur will call on the South African President at the Hotel Taj Palace at 11 a.m.
The half-an-hour meeting will be followed by delegation-level talks and a signing of bilateral agreements at Hyderabad House.
From 3 p.m. to 4.30 p.m., Dr. Zuma will address captains of Indian industry and expected to pitch South Africa as a favourable destination for Indian investment.
Leader of Opposition in the Lok Sabha, Sushma Swaraj will call on Dr. Zuma. He will then receive the Vice President of India, Mohammad Hamid Ansari. These two meetings will last for 30-minutes each at the Hotel Taj Palace.
Dr. Zuma will meet UPA Chairperson Sonia Gandhi at the same hotel at 5.30 p.m.
At 7.30 p.m. he will call on President Pratibha Devisingh Patil. The 30-minute meeting will be followed by a State Banquet hosted by the Indian President at Rashtrapati Bhawan.
Dr. Zuma and his entourage will emplane for South Africa at 10.30 p.m. (ANI)