Minimizing over a plane

Using Lagrange multipliers and define $F(x,y,z,\lambda) = x^2 + 2y^2 + 3z^2 + xy + 3yz + 2zx + 7x - \lambda (x+2y+4z - 10)$ . Taking partial derivatives and equating them to zero, we have:

$\begin{array} {lll} \dfrac {\partial F}{\partial x} = 2x + y + 2z - \lambda & \implies 2x + y + 2z = \lambda & ...(1) \\ \dfrac {\partial F}{\partial y} = x + 4y + 3z - 2\lambda & \implies x + 4y + 3z = 2 \lambda & ...(2) \\ \dfrac {\partial F}{\partial z} = 2x + 3y + 6z + 7 - 4\lambda & \implies 2x + 3y + 6z + 7 = 4\lambda & ...(3) \\ \dfrac {\partial F}{\partial \lambda} = - x - 2y - 4z + 10 & \implies x + 2y + 4z = 10 & ...(4) \end{array}$

From $2\times (2) - (3): \ 5y - 7 = 0 \implies y = \dfrac 75$ .

From $2 \times (1) - (2): \ 3x - 2y + z = 0 \ \ ...(5)$

From $4 \times (5) - (4): \ 11x - 10y = - 10 \implies 11 x = 10y - 10 \implies x = \dfrac 4{11}$

From $(5): \ \dfrac {12}{11} - \dfrac {14}5 + z = 0 \implies z = \dfrac {94}{55}$

Therefore $\min (F(x,y,z,0)) = F \left(\dfrac 4{11}, \dfrac 75, \dfrac {94}{55}, 0 \right) = \dfrac {1854}{55} \approx \boxed{33.7}$ . We can check that it is a minimum, since $F(10,0,0,0) = 100 > 33.7$ as $(x,y,z) = (10,0,0)$ satisfy $x+2y+4z = 10$