Minimize $x + 2y + 3z$ subject to $x^2 + y^2 + z^2 = 1$ and $x + y + z \le 0$ I'm trying to solve this problem by KKT's condition:

$$\begin{align*}
\text{min} & \quad x + 2y + 3z \\
\text{s.t} & \quad x^2 + y^2 + z^2 && = 1 \\
           & \quad x + y + z && \le 0
\end{align*}$$

The linear independence constraint qualification - LICQ is

The gradients of the active inequality constraints and the gradients of the equality constraints are linearly independent at $x^{*}$.

Could you please verify if I correctly apply the KKT's theorem? Thank you so much for your help!

$\textbf{My attempt:}$
Let $f = x + 2y + 3z$, $h = x^2+y^2 + z^2-1$, $g = x + y + z$, and $$\mathcal K= \{(x,y,z) \in \mathbb R^3 \mid h (x,y,z) = 0 \text{ and } g (x,y,z) \le 0\}$$
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 $\nabla g (x,y,z) = \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}$.
Consider the system $$\begin{cases}
 \mu_1 \nabla h (x,y,z) + \mu_2 \nabla g (x,y,z) &=0 \\
  h (x,y,z) &= 0\\
  g (x,y,z) &=0
\end{cases}
\iff 
\begin{cases}
 \mu_1  \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} + \mu_2 \begin{pmatrix} 1\\ 1 \\ 1 \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix} \\
  x^2 + y^2 + z^2 &= 1\\
  x   + y   + z   &= 0
\end{cases}$$
$$\iff \begin{cases}
2x \mu_1 + \mu_2 &= 0\\
2y \mu_1 + \mu_2 &= 0 \\
2z \mu_1 + \mu_2 &= 0 \\
  x^2 + y^2 + z^2 &= 1\\
  x   + y   + z   &= 0
\end{cases}\iff
\begin{cases}
2\mu_1(x+y+z) + 3\mu_2 &= 0\\
2x \mu_1 + \mu_2 &= 0 \\
2y \mu_1 + \mu_2 &= 0 \\
2z \mu_1 + \mu_2 &= 0 \\
  x^2 + y^2 + z^2 &= 1\\
  x   + y   + z   &= 0
\end{cases}$$
$$\iff
\begin{cases}
\mu_2 &= 0\\
x \mu_1 &= 0 \\
y \mu_1 &= 0 \\
z \mu_1 &= 0 \\
  x^2 + y^2 + z^2 &= 1\\
  x   + y   + z   &= 0
\end{cases} \implies \mu_1 = \mu_2 = 0$$
Hence LICQ is satisfied. It follows from KKT's theorem that the solution satisfies
$$\begin{cases}
\mu_2 &\ge 0\\
h (x,y,z) &= 0\\
\mu_2  g (x,y,z) &=0 \\
\nabla f (x,y,z) + \mu_{1} \nabla  h (x,y,z) + \mu_{2} \nabla  g (x,y,z)&=0
\end{cases}
\iff
\begin{cases}
  \mu_2 &\ge 0\\
  x^2+y^2 + z^2 &=1\\
  \mu_2 (x + y + z)   &= 0 \\
 \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} +\mu_1  \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} + \mu_2 \begin{pmatrix} 1\\ 1 \\ 1 \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix}
\end{cases}$$
$$\iff
\begin{cases}
  \mu_2 &\ge 0\\
  x^2+y^2 + z^2 &=1\\
  \mu_2 (x + y + z)   &= 0 \\
1 + 2x \mu_1 + \mu_2 &= 0\\
2 + 2y \mu_1 + \mu_2 &= 0 \\
3+ 2z \mu_1 + \mu_2 &= 0 \\
\end{cases}\iff\begin{cases}
 \mu_1 &= \sqrt 7 / \sqrt 2 \\
 \mu_2 &= 0 \\
 x  &= -1/\sqrt{14} \\
 y  &= -2/\sqrt{14} \\
 z  &= -3/\sqrt{14}
\end{cases}$$
As such, the solution is $(x,y,z)=(-1/\sqrt{14},-2/\sqrt{14},-3/\sqrt{14})$.

Let $k = x + 2y + 3z$ and observe that it represents a plane with $k$, $\frac k2$ and $\frac k3$ as the intercepts with the $x$, $y$ and $z$ axes, respectively. So, $k$ is at the minimum if the intercepts have the largest negative values, corresponding to the plane tangential to the sphere $x^2 + y^2 + z^2=1$ in the negative octant.
The normal vector to the plane is $n=(1, 2, 3)$ and the radius parallel to $n$ is $t(1,2,3)$, or 
$$x=t,\>\>\>\>\>y = 2t,\>\>\>\>\>z=3t$$
Plug the above radial line into the equation of the sphere to get $t=\pm \frac1{\sqrt{14}}$ and the tangent point for the minimum $k$ is
$$x_m=-\frac1{\sqrt{14}},\>\>\>\>\>y_m = -\frac2{\sqrt{14}},\>\>\>\>\>z_m=-\frac3{\sqrt{14}}$$
Thus, the minimum value is, 
$$k_m = x_m + 2y_m + 3z_m = -\sqrt{14}$$
