Answer
Minimum value of $f(x,y,z)=x^2+2y^2+3z^2$ is
$\dfrac{33}{23}$
Work Step by Step
$f(x,y,z)=x^2+2y^2+3z^2$; $x+y+z=1$;$x-y+2z=2$
$x=\frac{18}{23}$
$y=-\frac{6}{23}$
$z=\frac{11}{23}$
Minimum value of $f(x,y,z)=x^2+2y^2+3z^2$ is
$(\frac{18}{23})^2+2(-\frac{6}{23})^2+3(\frac{11}{23})^2=\frac{33}{23}$
