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?

An option:
1) $x+y+z=1$; and 
2) $3x+y+z=5$;
$2$ planes , their intersection is a straight line.
Subtract: 2)-1):
$2x=4$; $x=2$ ;and 
$y+z=-1$;
$d^2=x^2+y^2+z^2$ .
Minimal distance of line from origin:
$d^2= 4 +y^2+z^2.$
2D problem:
Minimize $y^2+z^2$ with constraint $y+z=-1$.
$d_2^2= $
$[-(1+z)]^2+z^2=2z^2+2z+1=$
$2(z^2+z)+1= $
$2[(z+1/2)^2]-1/2+1\ge 1/2$.
Equality at $z=-1/2$; 
Finally:
Minimum at :
$x=2$ ; $y=-1/2$; $z=-1/2$;
$d^2_{\min}= 4+1/2=9/2;$
