<s>[INST] Minimal distance between the origin and intersection of $x^2 = 2yz$ and $x^2+3y^2+2z^2 = 30$ 
Find the minimal distance between the origin and intersection of $x^2 = 2yz$ and $x^2+3y^2+2z^2 = 30$

Attempt:
$2yz+3y^2+2z^2 = 30$
I called $f(y,z) = 2yz+3y^2+2z^2$
Then, $\nabla f(y,z) = (2z+6y,2y+4z) = 0 \Rightarrow y=z=0$
So the critical point is $(0,0,0)$, which is clearly a minimum. But, I think this is wrong, because the minimum point is not necessarily the minimal distance to the origin. How do I get the closest point to the origin?
 [/INST] I'll try to answer my own question based on the hints.
Define $f(x,y,z)=x^2+y^2+z^2$, $\varphi_1(x,y,z)=x^2-2yz$ and $\varphi_2(x,y,z)=x^2+3y^2+2z^2$
By doing $\nabla f = \lambda_1\varphi_1+\lambda_2\varphi_2$, we have
$(2x,2y,2z)=\lambda_1(2x,-2z,-2y)+\lambda_2(2x,6y,4z)$
$2x = \lambda_12x+\lambda_22x$
$2y =-\lambda_12z+\lambda_26y$
$2z=-\lambda_12y+\lambda_24z$
Together with $\varphi_1 = 0$ and $\varphi_2=30$, calculating this system of equation, we have:
$y=(3/2)z$ and $x =\pm\sqrt3z$, so $z=\pm\sqrt{60/19}$
And the points are $(\pm\sqrt3\sqrt{60/19},(3/2)\sqrt{60/19},\sqrt{60/19})$ and $(\pm\sqrt3\sqrt{60/19},-(3/2)\sqrt{60/19},-\sqrt{60/19})$
which are all minimum points as $\det Hf(x,y,z) = 8>0$, as
$Hf(x,y,z)=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
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