det `[[1,1,1],[sinA,sinB,sinC],[sin^2A,sin^2B,sin^2C]]=`

With the usual notation in DeltaABC det |(1,1,1),(sinA,sinB,sinC),(sin^(2)A,sin^(2)B,sin^(2)C)| assumes the value

In a Delta ABC |[1,1,1],[1+sinA,1+sinB,1+sinC][sinA+sin^2A,sinB+sin^2B,sinC+sin^2C]|=0

In a triangleABC , if |[1,1,1][1+sinA,1+sinB,1+sinC],[sinA+sin^2A, sinB+sin^2B, sinC+sin^2C]|=0 , then prove that triangleABC is an isosceles triangle.

If A, B and C are the angles of a triangle and |(1,1,1),(1+sinA,1+sinB,1+sinC),(sinA+sin^2A,sinB+sin^2B,sinC+sin^2C)|=0 , then the triangle must be :

In a triangleABC , if |(1,1,1),(1+sinA,1+sinB,1+sinC),(sinA+sin^2A,sinB+sin^2B,sinC+sin^2C)|=0 , then prove that triangleABC is an isosceles triangle.

In a triangleABC , if |(1,1,1),(1+sinA,1+sinB,1+sinC),(sinA+sin^2A,sinB+sin^2B,sinC+sin^2C)|=0 , then prove that triangleABC is an isosceles triangle.

If A ,B and C are the angles of an equilateral triangle, then the value of |{:(1,1,1),(1+sinA,1+sinB,1+sinC),(sinA+sin^2A,sinB+sin^2B,sinC+sin^2C):}| is ...........