The line $L$ is passing through $(1,2,3)$. The distance of any point on the line $L$ from the line $\bar{r}=(3 \lambda-1) \hat{i}+(-2 \lambda+3) \hat{j}+(4+\lambda) \hat{k}$ is constant. Then the line $L$ does not pass through the point

Given line $L'$ is parameterized as $\vec{r} = (-\hat{i} + 3\hat{j} + 4\hat{k}) + \lambda(3\hat{i} - 2\hat{j} + \hat{k})$. It passes through $A(-1, 3, 4)$ with direction vector $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.

The distance from any point on line $L$ to $L'$ is constant, implying $L$ is parallel to $L'$. Thus, $L$ shares direction vector $\vec{d} = \vec{b}$ and passes through $P(1, 2, 3)$, giving its parametric form as $(1 + 3k, 2 - 2k, 3 + k)$ where $k$ is a scalar.

For a point to lie on $L$, the ratios $\frac{x-1}{3} = \frac{y-2}{-2} = \frac{z-3}{1}$ must be equal to some $k$.

Points $(4, 0, 4)$, $(-2, 4, 2)$, and $(7, -2, 5)$ satisfy this condition with $k = 1$, $k = -1$, and $k = 2$ respectively.
Point $(-5, 6, 2)$ yields inconsistent values: $\frac{-5-1}{3} = -2$, $\frac{6-2}{-2} = -2$, $\frac{2-3}{1} = -1$, so it does not lie on $L$.

The point not on line $L$ is $\boxed{D}$.