<s>[INST] Find any local max or min of $x^2+y^2+z^2$ s.t $x+y+z=1$ and $3x+y+z=5$ 
Find any local max or min of
  \begin{align}
f(x,y,z)=x^2+y^2+z^2 && (1)
\end{align}
  such that 
  \begin{align}
x+y+z=1 && (2)\\
3x+y+z=5 && (3)
\end{align}

My attempt. Let
$L(x,y,z,\lambda_1, \lambda_2)= f(x,y,z)+\lambda_2 (x+y+z-1) + \lambda_1 (3x+y+z-5)$
$L_x=2x+ 3 \lambda_1 + \lambda_2 =0$ 
$L_y=2y+ \lambda_1 + \lambda_2=0$
$L_z=2z+\lambda_1 + \lambda_2=0$ 
Solve for $x,y,z$ we get:
$x=\frac{-3 \lambda_1 - \lambda_2}{2}$
$z=y=\frac{-\lambda_1 - \lambda_2}{2}$
with the use of $(2)$ and $(3)$ $\implies$
$x=2$
$y=z= \frac{-1}{2}$
so the stationary point is $(x,y,z)=(2, \frac{-1}{2},\frac{-1}{2})$
The Hessian of $L$ gives a postive definite matrix for all $(x,y,z)$ thus
$(x,y,z)=(2, \frac{-1}{2},\frac{-1}{2})$ is the only local minimizer of $f$ and there is no maximizors of $f$.
Is this correct?
 [/INST] Your result is correct. This is an alternative approach. 
Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $t\in \mathbb{R}$.
Hence we reduce the problem to a $1$-variable case,
$$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$
The above quadratic polynomial has just one local/global minimum at $t=-1/2$ that is at $(2,-1/2,-1/2)$. There is no local/global maximum.
</s>