I assume there are some plus signs that aren't rendering for some reason, so that the plane should be \(x+y+z=1\).
You're minimizing \(d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}\) subject to the constraint \(f(x,y,z)=x+y+z=1\). Note that \(d(x,y,z)\) and \(d(x,y,z)^2\) attain their extrema at the same values of \(x,y,z\), so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.
The Lagrangian is
\(L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)\)
Take your partial derivatives and set them equal to 0:
\(\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}\)
Adding the first three equations together yields
\(2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43\)
and plugging this into the first three equations, you find a critical point at \((x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)\).
The squared distance is then \(d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43\), which means the shortest distance must be \(\sqrt{\dfrac43}=\dfrac2{\sqrt3}\).
