My professor gave us this problem
$$\begin{align*} \text{min} & \quad x + 2y + 3z \\ \text{s.t} & \quad x^2 + y^2 + z^2 && = 1 \\ & \quad x + y + z && \le 0 \end{align*}$$
and a hint
Define $$\begin{aligned}f&:\mathbb R^3 \to \mathbb R, \quad (x,y,z) \mapsto x + 2y + 3z \\ h &:\mathbb R^3 \to \mathbb R, \quad (x,y,z) \mapsto x^2+y^2 + z^2-1 \\ g &:\mathbb R^3 \to \mathbb R, \quad (x,y,z) \mapsto x + y + z\end{aligned}$$ and $$\mathcal K= \left \{(x,y,z) \in \mathbb R^3 \,\middle\vert\, \begin{aligned} h (x,y,z) = 0 \\ g (x,y,z) \le 0\end{aligned} \right\}$$
Because $\mathcal K$ is compact and $f$ is continuous, the problem has a solution.
Moreover, $\nabla f (x,y,z) = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$, $\nabla h (x,y,z) = \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}$, and $\nabla g (x,y,z) = \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}$.
Use the linearly independent family and $v = \begin{pmatrix} -1 \\ -1 \\ -1 \\ \end{pmatrix}$. Then the constraints are qualified for any point in $\mathcal K$.
The hint leads me to Mangasarian-Fromovitz constraint qualification - MFCQ, which is
The gradients of the equality constraints are linearly independent at $x^{*}$ and there exists a vector $v \in \mathbb{R}^{n}$ such that $\langle \nabla g_{i}\left(x^{*}\right), v \rangle<0$ for all active inequality constraints and $\langle \nabla h_{j}\left(x^{*}\right), v\rangle=0$ for all equality constraints.
I have tried to finish this exercise but fail to follow his hint.
My attempt:
Consider $$\begin{cases} \mu \nabla h (x,y,z) &= 0 \\ h(x,y,z) &= 0 \end{cases} \iff \begin{cases} \mu \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ x^2 + y^2 + z^2 &= 1 \end{cases} \implies \mu =0$$
Hence the gradients of the equality constraints are linearly independent at the solution.
For $v = \begin{pmatrix} -1 \\ -1 \\ -1 \\ \end{pmatrix}$, we have $$\begin{cases} h(x,y,z) &= 0 \\ g(x,y,z) &= 0 \end{cases} \implies \langle \nabla g (x,y,z),v \rangle = \left \langle \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix},\begin{pmatrix} -1 \\ -1 \\ -1 \\ \end{pmatrix} \right \rangle = -3 <0$$
But $$\begin{cases} h(x,y,z) &= 0 \\ g(x,y,z) &\le 0 \end{cases} \implies\langle \nabla h (x,y,z),v \rangle = \left \langle \begin{pmatrix} 2x \\ 2y \\ 2z \\ \end{pmatrix},\begin{pmatrix} -1 \\ -1 \\ -1 \\ \end{pmatrix} \right \rangle =-2(x+y+z) \ge 0$$
whereas our desired result is $$\begin{cases} h(x,y,z) &= 0 \\ g(x,y,z) &\le 0 \end{cases} \implies\langle \nabla h (x,y,z),v \rangle = \left \langle \begin{pmatrix} 2x \\ 2y \\ 2z \\ \end{pmatrix},\begin{pmatrix} -1 \\ -1 \\ -1 \\ \end{pmatrix} \right \rangle = 0$$
Could you please elaborate on the correctness of this hint and how to choose a vector $v$ to finish this exercise?