The minimum possible perimeter of a triangle with one vertex at $\left(3,9\right)$ , one anywhere on the$y-$axis and one anywhere on the line $y=x$ is $\sqrt{K}(K\in N)$ Sum of digits of $k=$  ________

Let A be the vertex (3, 9), B be the vertex on the y-axis and C be the vertex on the line y
= x. Also, let D (-3, 9) be the reflection of A in the y-axis and E(9, 3) be the reflection of
A in the line y = x. Then AB = BD and AC = CE, and thus the perimeter of ABC is
equal to DB + BC + CE.
But the shortest distance between two points is a straight line, so
$\begin{array}{r}\mathrm{DB}+\mathrm{BC}+\mathrm{CE}\ge \mathrm{DE}\\ =\sqrt{(9-(-\beta ){)}^2+(3-9{)}^2}=\sqrt{180}=6\sqrt{5}\end{array}$
This minimum can be obtained by then choosing B and C as the
points of intersection of the line DE with the y-axis and the line y = x, respectively. This
gives us the points $B\left(0,\frac{15}{2}\right)\text{ and }C(5,5)$This will yield a perimeter for $\mathrm{△}ABC\text{ of }6\sqrt{5}\text{. }$