Minimization of sphere in 3D with two linear constraints

Question

I am trying to solve the minimization problem below. In the first case with only one constraint, the optimal point is easily reached to be $x* = (1,1,1)$ with Lagrange's multiplier value $λ=-1$. But, in the second case with one additional constraint, the new constraint is more extended than the first one and consequently I get the answer $x* = (6/7,12/7,17/7)$. Thus, the optimal point is again $x* = (1,1,1)$ which comes from a tighter constraint, isn't it?

*Black plane: $x_{1} + x_{2} + x_{3} = 3$

**Blue Plane: $x_{1} + 2x_{2} + 3x_{3} = 12$

Then, why the problem statement mentions that "the second minimum value should be less than the first minimum value"??

In other words, how the second constraint $x_{1} + 2x_{2} + 3x_{3} = 12$ really affect the previous optimal point reached by applying the first constraint $x_{1} + x_{2} + x_{3} = 3$?

Find the minimum value of $F(x) = 1/2 (x_{1} + x_{2} + x_{3})$ with one constraint $x_{1} + x_{2} + x_{3} = 3$ and then with an additional constraint $x_{1} + 2x_{2} + 3x_{3} = 12$. The second minimum value should be less than the first minimum value : Why ? The first problem has a ??? tangent to a sphere in R3 . The second problem has a ??? tangent to a sphere in R3 .

Answer 1

Think that if the restrictions separated are

$$ \cases{ R_1\to x_1+x_2+x_3 = 3\\ R_2\to x_1+2x_2+3x_3 = 12} $$

the minimization is with $R = R_1\cap R_2$ or the restriction $R$ defines a line while each of the restrictions $R_1,R_2$ define planes. Obviously $R\in R_1$ and $R\in R_2$ so consequently

$$ \min_{x\in R} F(x) \ge \min_{x\in R_1} F(x) $$

or

$$ \min_{x\in R} F(x) \ge \min_{x\in R_2} F(x) $$

Here we assumed $F(x) = \frac 12(x_1^2+x_2^2+x_3^2)$

NOTE

In the last case, the lagramgian reads

$$ L(x,\lambda) = F(x) + \lambda_1 R_1 +\lambda_2 R_2 $$

and the stationary points are given by the solutions to

$$ \nabla L = \cases{x_1+\lambda_1 + \lambda_2 = 0\\ x_2 +\lambda_1 + 2\lambda_2 = 0 \\ x_3 + \lambda_1+3\lambda_3 = 0\\ x_1+x_2+x_3 = 3 \\ x_1+2x_2+3x_3 = 12} $$

giving only one stationary point which is $x^* = (-2,1,4)$ with $F(x^*) = 10.5$