Yes, it is deal with vector spaces over real number. For a vector to be subspaces, it must be closed under addition and scalar multiplication.
(a) (a1, b1) + (a2, b2) = (a1 + a2, b1 + b2) => closed under addition
k(a, b) = (ka, kb) => closed under scalar multiplication
But i do think that if a>= 0, then k(a, b) = (ka, kb) is not closed under scalar multiplication when k is negative value so it is not subspace of V.
(b) (a1, c1, c1) + (a2, c2, c2) = (a1 + a2, c1 + c2, c1 + c2) => closed under addition
k(a, c, c) = (ka, kc, kc) => closed under scalar multiplication
Similarly, a^2 + b^2 + c^2 <= 1 could mean a, b, c might be 0, 0<= a <= 1 or -1 <= a <= 0. But i do not know why b is there since this question does not have value b.
and if a is 1, then a1 + a2 = 2 which is not closed under addition so it is not subspace too.
I hope that my working is correct. But i doubt that i explain it in correct way. Or maybe i shall prove that with contradiction.
(1, 0, 0) + (1, 0, 0) = (2, 0, 0) which is not closed under addition.
Could you further elaborate on that?
The claim "W is closed under scalar multiplication" means "For all real numbers k and all vectors (a, b, c) ∈ W, (ka, kb, kc) ∈ W." To form its negation, switch the quantifiers and add the negation to the final part: "There exists a real number k and a vector (a, b, c) ∈ W such that (ka, kb, kc) ∉ W." Can you find specific k, a, b, c for which this negation is true?
it appears there is a typo in the second part of the problem, and i believe it should be triples of the form (a,b,c) instead of (a,c,c).
i also suggest using a k much larger than 1 as a scalar for a counter-example for the second part.