Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$ , and suppose this triangle is acute (so all altitudes are on the interior of the triangle). Let the altitude to the side of length $x$ be of length $h_x$ , and similarly for $y$ and $z$ . Then we have by two applications of the Pythagorean Theorem that $x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}$ . As a function of $h_x$ , the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that $h_x^2 = \dfrac1{16}$ and so $h_x = \dfrac{1}4$ and similarly $h_y = \dfrac1{5}$ and $h_z = \dfrac1{6}$ Since the area of the triangle must be the same no matter how we measure, $x\cdot h_x = y\cdot h_y = z \cdot h_z$ and so $\dfrac x4 = \dfrac y5 = \dfrac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$ . The semiperimeter of the triangle is $s = \dfrac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have $A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}$ . Thus $A = \frac{1}{15\sqrt{7}}$ and $x + y + z = 30A = \frac2{\sqrt{7}}$ and the answer is $2 + 7 = \boxed{9}$

to get the answer follow the steps given below:

take one radical to LHS and square both sides in the 3 equations and simplify

now leave these 3 equations and take other radical to LHS (which was left in previous step) and square both sides.

get relation b/w x,y and x,z ie x=+-0.8*y and x=+-0.67z

put all values in terms of x in 1st equation and get x,y&z.

x+y+z=2/7^0.5

answer=9