I started with the equation $(x-2)^2+y^2+(z+3)^2$ as the main function, using the plane as the constraint function. From that I got
$2(x-2)=1*\lambda$, $2y=1*\lambda$, and $2(z+3)=1*\lambda$
and tried to solved for lambda making them all equal and solving for that but didn't get anywhere with that approach
$2x-4=\lambda$,$2y=\lambda$, $2z+6=\lambda$
and without knowing anyway to get rid of the lambda, I can't just try solving for any of the variables, so I'm not sure what else to try.
Solve the three equations for $x,y,z$, then plug them back into your constraint, which is what you still have to use.
You get $x=\frac{\lambda+4}2,y=\frac\lambda2,z=\frac{\lambda-6}{2}$. Hence $$1=x+y+z=x=\frac{\lambda+4}2+\frac\lambda2+\frac{\lambda-6}{2}.$$ Solving for $\lambda$ gives $$\lambda=\frac43.$$ Plugging it back into the previous equations gives you the values $$x=\frac83,y=\frac23,z=-\frac73.$$