If asinA +bsinB + csinC=k , then what is the minimum value of of sin^2(A) + sin^2(B) + sin^2(C)

The answer is k^2 / (a^2 + b^2 + c^2). It follows from the Cauchy Schwarz inequality. Let t be the sum sin^2(A) + sin^2(B) + sin^2(C). Then, by the Cauchy-Schwarz inequality,

t (a^2 + b^2 + c^2) >= (asinA +bsinB + csinC)^2 = k^2

So t >= k^2 / (a^2 + b^2 + c^2).

Equality holds when a/sin A = b/sin B = c/sin C, i.e., A, B, C are the angles opposite the sides a,b,c in a triangle.

let x =sin(A), y =sin(B), z =sin(C) here

so have a plane: ax + by +cz =k
where the distance function k/√(x² + y² + z²} is minimal,
this occurs when vector (x,y,z) is perpendicular to plane.

the director for the plane normal is vector (a,b,c)
minimum value required is k²/(a² + b² + c²)

e.g. 3sin(A) + 4sin(B) + 5sin(C) = 10 gives sum = 100/50 = 2