a) Para provarmos que o conjunto é espaço vetorial, realizaremos os cálculos abaixo:

\(\begin{align} & x={{z}^{2}} \\ & u=({{x}_{1}},{{y}_{1}},{{z}_{1}})=(z_{1}^{2},0,{{z}_{1}}) \\ & v=({{x}_{2}},{{y}_{2}},{{z}_{2}})=(z_{2}^{2},0,{{z}_{1}}) \\ & \\ & u+v=(z_{1}^{2},0,{{z}_{1}})+(z_{2}^{2},0,{{z}_{2}}) \\ & u+v=\left( z_{1}^{2}+z_{2}^{2},0,{{z}_{1}}+{{z}_{2}} \right) \\ & \\ & \alpha u=\alpha ({{x}_{1}},{{y}_{1}},{{z}_{1}}) \\ & \alpha u=\alpha (z_{1}^{2},0,{{z}_{1}}) \\ & \alpha u=(\alpha z_{1}^{2},0,\alpha {{z}_{1}}) \\ \end{align}\ \)

\(\boxed{\alpha u = \left( {\alpha z_1^2,0,\alpha {z_1}} \right)}\)

b)

\(\begin{align} & x+y-2z=0 \\ & u=({{x}_{1}},{{y}_{1}},{{z}_{1}})=(x_{1}^{{}},{{y}_{1}},-2{{z}_{1}}) \\ & v=({{x}_{2}},{{y}_{2}},{{z}_{2}})=(x_{2}^{{}},{{y}_{2}},-2{{z}_{2}}) \\ & \\ & u+v=(x_{1}^{{}},{{y}_{1}},-2{{z}_{1}})+(x_{2}^{{}},{{y}_{2}},-2{{z}_{2}}) \\ & u+v=(x_{1}^{{}}+{{x}_{2}},{{y}_{1}}+{{y}_{2}},-2{{z}_{1}}-2{{z}_{2}}) \\ & \\ & \alpha u=\alpha ({{x}_{1}},{{y}_{1}},-2{{z}_{1}}) \\ & \alpha u=(\alpha {{x}_{1}},\alpha {{y}_{1}},-\alpha 2{{z}_{1}}) \\ \end{align}\ \)

\(\boxed{\alpha u = \left( {\alpha x_1^{},\alpha {y_1}, - 2\alpha {z_1}} \right)}\)