Inspired by Steven Chase

The equation of the disk can be parametrized as: $\vec{r}(R,\theta)=\vec{C}+R\cos\theta \hat{e_1} + R\sin\theta \hat{e_2}$ where $0 \leq R \leq 4$ , $0 \leq \theta \leq 2\pi$ and $\{\hat{e_1},\hat{e_2}\}$ is an orthonormal basis for the plane in which the disk is contained. With a little luck we find that $(1,-1,1)$ belongs to the plane $-2x+5y+7z=0$ , so $(1,-1,1) \times \vec{N} = (4,3,-1)$ will be the other spanning vector. So we just need to normalize: $\hat{e_1}=\dfrac{1}{\sqrt{3}}(1,-1,1)$ and $\hat{e_2}=\dfrac{1}{\sqrt{26}}(4,3,-1)$ .

Let $f(R,\theta)=\lVert \vec{r} - (5,7,9) \rVert^2$ , then: $\begin{aligned} f(R,\theta) &= \lVert (-2,-6,-4)+R\cos\theta \hat{e_1} + R\sin\theta \hat{e_2} \rVert^2 \\ &= \lVert (-2,-6,-4) \rVert^2 + R^2\cos^2\theta \lVert \hat{e_1} \rVert^2 + R^2\sin^2\theta \lVert \hat{e_2} \rVert^2 + 2R\cos\theta(-2,-6,-4) \cdot \hat{e_1} + 2R\sin\theta(-2,-6,-4) \cdot \hat{e_2} + R^2\cos\theta\sin\theta\hat{e_1} \cdot \hat{e_2} \\ &= 56 + R^2 - \dfrac{44}{\sqrt{26}}R\sin\theta \end{aligned}$ Now, to find the minimum distance, find $(R,\theta)$ such that $\nabla f = \vec{0}$ : $\begin{aligned} 0 &= \dfrac{\partial f}{\partial \theta} = -\dfrac{44}{\sqrt{26}}R\cos\theta \implies \theta = \dfrac{\pi}{2}\\ 0 &= \dfrac{\partial f}{\partial R} = 2R - \dfrac{44}{\sqrt{26}}\sin\theta \implies R = \dfrac{22}{\sqrt{26}}\sin\theta = \dfrac{22}{\sqrt{26}} \end{aligned}$ But $\dfrac{22}{\sqrt{26}} > 4$ , so we are outside the disk. But since the distance is decreasing from the center of the disk until $R=\dfrac{22}{\sqrt{26}}$ keeping fixed $\theta=\dfrac{\pi}{2}$ , we must choose $R=4$ , i.e., a point on the boundary: $\begin{aligned} f_{min} &= f\left(4,\dfrac{\pi}{2}\right) = 56 + 16 - \dfrac{44}{\sqrt{26}}(4)\sin \dfrac{\pi}{2} \\ &= 72 - \dfrac{176}{\sqrt{26}} = 72 - 88\sqrt{\dfrac{2}{13}} \end{aligned}$

Finally, the minimum distance is just the square root of the previous value: $\sqrt{72 - 176\sqrt{\dfrac{1}{26}}}$ , making the answer $72+176+26=\boxed{274}$ .

Solution without using Calculus:

Let $\vec{u}$ be a vector from the center of the disc $(3, 1, 5)$ to an arbitrary point of the disc. Then $|\vec{u}|\leq 4$ and $\vec{u} \cdot \vec{N}=0 .$ Let $\vec{v}$ be the vector from the center of the disc $(3, 1, 5)$ to the point $(5, 7, 9).$ Then our problem reduces to minimize $|\vec{v}-\vec{u}|$ for all possible vectors $\vec{u}$ defined as above. In what follows, we are going to use that the vector $\vec{v}$ can be expressed in the form $\vec{v}=\vec{v_1}+\vec{v_2},$ where $\vec{v_1}=\frac{\vec{v}\cdot\vec{N}}{|\vec{N}|^2} \vec{N}$ -that is, parallel to $\vec{N}-$ and where $\vec{v_2}\cdot\vec{v_1}=0.$

Let's find an expression for $|\vec{v}-\vec{u}|^2.$ $|\vec{v}-\vec{u}|^2=(\vec{v}-\vec{u})\cdot(\vec{v}-\vec{u})=|\vec{v}|^2-2\vec{v}\cdot\vec{u}+|\vec{u}|^2=56+|\vec{u}|^2-2\vec{u}\cdot(\vec{v_1}+\vec{v_2}).$ Since $\vec{u}$ and $\vec{v_1}$ are orthogonal then $|\vec{v}-\vec{u}|^2=56+|\vec{u}|^2-2\vec{u}\cdot\vec{v_2}=56+|\vec{u}|^2-2|\vec{u}||\vec{v_2}|\cos \theta\geq 56+|\vec{u}|^2-2|\vec{u}||\vec{v_2}|=56+(|\vec{u}|-|\vec{v_2}|)^2-|\vec{v_2}|^2\quad\quad(*)$ where $\theta$ is the angle formed by $\vec{u}$ and $\vec{v_2}.$ Now, using the fact that $\vec{v}=\vec{v_1}+\vec{v_2},$ and since $\vec{v_2}\cdot\vec{v_1}=0.$ we can conclude that $|\vec{v}|^2=|\vec{v_1}|^2+|\vec{v_2}|^2,$ and, therefore, $|\vec{v_2}|=\frac{22}{\sqrt{26}},$ and this value is greater than 4. Using this fact, the fact that $|\vec{u}|\leq 4$ , and $(*),$ $|\vec{v}-\vec{u}|^2 \geq 56+(4-|\vec{v_2}|)^2-|\vec{v_2}|^2=72-\frac{176}{\sqrt{26}}$ The equality is reached when $\vec{u}=4\frac{\vec{v_2}}{|\vec{v_2}|}.$ Therefore, the minimum value of $|\vec{v}-\vec{u}|$ will be $\sqrt{72-\frac{176}{\sqrt{26}}}.$ So the answer to the question will be $72+176+26=\boxed{274}.$