Answer
$\frac{{25}}{3}$ is the minimum value of $f$ subject to the two constraints.
Work Step by Step
In this exercise our task is to minimize the function $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ subject to the two constraints $g\left( {x,y,z} \right) = x + y + z - 1 = 0$ and $h\left( {x,y,z} \right) = x + 2y + 3z - 6 = 0$.
The Lagrange condition is
$\nabla f = \lambda \nabla g + \mu \nabla h$
$\left( {2x,2y,2z} \right) = \lambda \left( {1,1,1} \right) + \mu \left( {1,2,3} \right)$
From the Lagrange condition we obtain three equations:
$2x = \lambda + \mu $, ${\ \ }$ $2y = \lambda + 2\mu $, ${\ \ }$ $2z = \lambda + 3\mu $
Subtracting the second equation from the first equation gives
$2x - 2y = - \mu $
(1) ${\ \ \ \ }$ $\mu = 2y - 2x$
Subtracting the third equation from the first equation gives
$2x - 2z = - 2\mu $
(2) ${\ \ \ \ }$ $\mu = z - x$
Subtracting the third equation from the second equation gives
$2y - 2z = - \mu $
(3) ${\ \ \ \ }$ $\mu = 2z - 2y$
Equations (1), (2) and (3) gives
$2y - 2x = z - x = 2z - 2y$
$2y - 2x = z - x$
So, $z=2y-x$.
Substituting $z=2y-x$ in $g\left( {x,y,z} \right)$ and $h\left( {x,y,z} \right)$ yield
$x + y + 2y - x - 1 = 0$, ${\ \ }$ $x + 2y + 6y - 3x - 6 = 0$
$3y = 1$, ${\ \ \ }$ $ - 2x + 8y - 6 = 0$
So, $y = \frac{1}{3}$.
Substituting $y = \frac{1}{3}$ in the second equation above gives $x = - \frac{5}{3}$.
Using $z=2y-x$, we get $z = \frac{7}{3}$.
So, the critical point is $\left( { - \frac{5}{3},\frac{1}{3},\frac{7}{3}} \right)$.
The extreme value corresponds to this critical point is $f\left( { - \frac{5}{3},\frac{1}{3},\frac{7}{3}} \right) = \frac{{25}}{3}$.
Since the function $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ is increasing, we conclude that $\frac{{25}}{3}$ is the minimum value of $f$ subject to the two constraints. This is illustrated in the figure attached.
