Find the shortest distance from the origin to the surface of $x=yz+10$ we can use the distance formula, but for the sake of simplicity we can use $f(x,y,z)=x^2+y^2+z^2$  and let $\phi(x,y,z)=x-yz=10$. Via Lagrange multiplier, 
\begin{align}\partial_xf+\partial_x\phi\cdot\lambda=0\\
2x+\lambda=0\tag{i}\\
\partial_yf+\partial_y\phi\cdot\lambda=0\\
2y-z\lambda=0\tag{ii}\\
\partial_zf+\partial_z\phi\cdot\lambda=0\\
2z-\lambda y=0\tag{iii}
\end{align}
from equating (ii) and (iii) we find that $$2z=\frac{\lambda^2z}{2}\implies \lambda=\pm 2$$ 
for $\lambda=2,y=z,$ and $x=-1$, which means 
$$-1-yz=10$$
$$-yz=11$$
but since $y=z$
$$y^2=-11$$
Therefore $\lambda=-2\implies y=-z,x=1$
\begin{align}
-y^2=-9\\
y=3\\
z=-3 
\end{align}
Therefore the shortest distance from the origin to the plane is $\sqrt{9+9+1}=\sqrt{19}$
I believe this should be correct, but I have no idea how to verify it, could anyone help me confirm this, or point out where I made a mistake? Thanks!

We substitute $x = 10 - yz$ into squared distance $d^{2} = x^{2}+y^{2}+z^{2}$ to obtain $ d^{2}=100+20yz+(yz)^{2}+y^{2}+z^{2}$. Rewrite the equation as $d^{2}=19+(y+z)^{2}+(yz+9)^{2}$. This squared distance is minimized if $y+z=0$ and $yz+9=0$, the minimum squared distance is then $19$ and the minimum distance is $\sqrt{19}$.
