The coordinates of the foot of the perpendicular, denoted as \( P(x, y, z) \), are defined by the relation:\[\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} = \lambda.\]This yields the parametric form of the line:\[(x, y, z) = (4\lambda - 2, 2\lambda + 1, 3\lambda - 1).\]The vector \( \overrightarrow{AP} \) from point \( A(1, 2, 4) \) to \( P \) is calculated as:\[\overrightarrow{AP} = (4\lambda - 3) \hat{i} + (2\lambda - 1) \hat{j} + (3\lambda - 5) \hat{k}.\]The direction vector of the line is:\[\overrightarrow{b} = 4\hat{i} + 2\hat{j} + 3\hat{k}.\]Given that \( \overrightarrow{AP} \) is perpendicular to \( \overrightarrow{b} \), their dot product is zero:\[\overrightarrow{AP} \cdot \overrightarrow{b} = 0.\]Substituting the expressions for \( \overrightarrow{AP} \) and \( \overrightarrow{b} \) results in:\[4(4\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 5) = 0.\]Simplifying the equation gives:\[16\lambda - 12 + 4\lambda - 2 + 9\lambda - 15 = 0,\]which reduces to:\[29\lambda - 29 = 0.\]Solving for \( \lambda \):\[\lambda = 1.\]Substituting \( \lambda = 1 \) back into the parametric equations of the line determines the coordinates of \( P \):\[P(2, 3, 2).\]The perpendicular distance from point \( P(2, 3, 2) \) to the plane \( 3x + 4y + 12z + 23 = 0 \) is computed using the distance formula:\[\text{Distance} = \frac{|3(2) + 4(3) + 12(2) + 23|}{\sqrt{3^2 + 4^2 + 12^2}}.\]The numerator simplifies to:\[|6 + 12 + 24 + 23| = 65.\]The denominator simplifies to:\[\sqrt{9 + 16 + 144} = \sqrt{169} = 13.\]Therefore, the perpendicular distance is:\[\text{Distance} = \frac{65}{13} = 5.\] Final Answer:\[\boxed{5}\]