Solution / Explanation:The vectors in the plane are $\vec{AB} = \langle -4, -8, 8 \rangle$ and $\vec{AC} = \langle 3, -2, 0 \rangle$. Their cross product is $\vec{AB} \times \vec{AC} = 16\hat{i} + 24\hat{j} + 32\hat{k}$, giving normal ratios $\langle 2, 3, 4 \rangle$. The plane equation through $(2, 5, -3)$ is $2x + 3y + 4z - 7 = 0$. The perpendicular distance from $(7, 2, 4)$ is $\frac{|2(7) + 3(2) + 4(4) - 7|}{\sqrt{2^2 + 3^2 + 4^2}} = \frac{29}{\sqrt{29}} = \sqrt{29}$.