The problem : Find the equation of the plane that passes through the points $(1,2,3)$ and $(3,2,1)$ and is perpendicular to the plane $4x-y+2z=7$
Solution : The given plane has a normal vector $\bar n_1$ perpendicular to itself and parallel to the second plane $$\bar n_1=4i-j+2k$$
The second plane, similarly, has a normal vector $\bar n_2$ which is the cross-product $$\bar n_2 = \bar n_1\times \bar{p_1p_2}$$
$\bar{p_1p_2}$ is the vector passing through points $(1,2,3)$ and $(3,2,1)$. The cross product is the determinant of a $3\times 3$ matrix where the first row consists of the unit vectors $i,j,k$, the second of $\bar n_1$ and the third of $\bar{p_1p_2}$ $$\bar n_2 = 4i+4j-6k$$
And the equation of the given plane through point $(1,2,3)$ $$4x+4y-6z=-6$$
First of all, I don't understand what the significance of =7 in the equation of the first plane has to do with the problem. I know we can find the $x,y,z$ intercepts with that, but the normal vector should be the same irregardless of the RHS. Secondly, I don't necessarily want to know whether this is correct, but rather how to go about proving that it is. Proving the normal vectors $\bar n_1$ and $\bar n_2$ are perpendicular to each other is easy - their dot product must be zero - but I don't think this alone is suffcient proof.
Your calculated value of ${\vec n_2}$ is wrong and should be $2{\hat i}+12{\hat j}+2{\hat k}$.
The $7$ in the original question is not very useful and can be ignored for the purpose of solving this question. The value could be anything and the normal vector will still be parallel to the required plane. To verify your answer, you can dot the normal vectors of both planes and verify that it is zero. You could also check that both points satisfy the given plane.