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