Yes, the set of vectors
V = {(x, y, z) : x - 2y + 3z = 0}
is indeed a vector space.
Let u = (x, y, z) and v = (r, s, t) be any two vectors in V. Then
x - 2y + 3z = 0
and
r - 2s + 3t = 0
Their vector sum is
u + v = (x + r, y + s, z + t)
We need to show that u + v also belongs to V - in other words, V is closed under summation. This is a matter of showing that the coordinates of u + v satisfy the condition on all vectors of V:
(x + r) - 2 (y + s) + 3 (s + t) = (x - 2y + 3z) + (r - 2s + 3t) = 0 + 0 = 0
Then V is indeed closed under summation.
Scaling any vector v by a constant c gives
cv = (cx, cy, cz)
We also need to show that cv belongs to V - that V is closed under scalar multiplication. We have
cx - 2cy + 3cz = c (x - 2y + 3z) = 0c = 0
so V is need closed under scalar multiplication.
