Closest point on a plane

The shortest path from a point to a plane is along the normal to the plane that passes through the given point. As $<3,-4,5>$ is normal to the plane, the point on the plane closest to $(1,2,3)$ will be of the form $(1,2,3) + (3,-4,5)t$ , where $t$ is such that this point satisfies the plane equation:

$3(1 + 3t) -4(2 - 4t) + 5(3 + 5t) = 30$

$\Longrightarrow 3 + 9t - 8 + 16t + 15 + 25t = 30 \Longrightarrow 50t = 20 \Longrightarrow t = \dfrac{2}{5}$ .

The closest point is then $\left(1 + \dfrac{6}{5}, 2 - \dfrac{8}{5}, 3 + 2\right) = \left(\dfrac{11}{5}, \dfrac{2}{5}, 5\right)$ , and thus

$5(a + b + c) = 5\left(\dfrac{11}{5} + \dfrac{2}{5} + \dfrac{25}{5}\right) = \boxed{38}$ .