Minimally fascinating

We can rewrite the given expression as ${ (x-1) }^{ 2 }+{ (2y-3) }^{ 2 }+3{ (z-1) }^{ 2 }+1$ , so the minimum value is $1$ , when $x=1$ , $y=3/2$ and $z=1$ .

Relevant wiki: Hessian Matrix

Let $f(x,y,z)=x^2+4y^2+3z^2-2x-12y-6z+14$

We have $\begin{cases} \frac{\partial{f}}{\partial{x}}=2x-2 \\ \frac{\partial{f}}{\partial{y}} = 8y-12 \\ \frac{\partial{f}}{\partial{z}} = 6z-6\end{cases}$

Solving we get $(x,y,z)=(1,1.5,1)$ . Indeed it's a stationary point of the function $f(x,y,z)$ . But the crucial part is we don't know exactly that is this a minima,maxima or saddle point even.

For this very part We use the Hessian to determine the nature of the point.

The Hessian Matrix of $f(x,y,z)$ is :

$\large \Delta^2f(x,y,z) = \begin{pmatrix} 2&0&0 \\ 0&8&0 \\ 0&0&6 \end{pmatrix}$ .

We readily observe it's a Positive definite matrix so $f$ attains a minimum & it exists.

So, $\boxed{f(x,y,x)\ge f(1,1.5,1)=1}$