Let the east direction be along the positive x-axis ($\hat{i}$), north be along the positive y-axis ($\hat{j}$), and vertically upward be along the positive z-axis ($\hat{k}$). The displacement vector of the body is $\vec{s} = 4\hat{i} + 3\hat{j} + 12\hat{k}$. The magnitude of this displacement vector is $|\vec{s}| = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13\text{ m}$. The option $19\text{ m}$ is the simple distance path length ($4+3+12=19$), which is incorrect for displacement. $5\text{ m}$ only considers the horizontal displacement ($4\hat{i} + 3\hat{j}$), neglecting the height. Thus, $13\text{ m}$ is the correct total displacement magnitude.