It's All Geometry!

Let $\overline{\rm AC} = 2x \implies \overline{\rm AB} = 3x$ and $\alpha + \lambda = 120^{\circ} \implies \alpha = 120^{\circ} - \lambda$ .

Using the law of cosines on $\triangle{ABC}$ with included $\angle{BAC} \implies y = \sqrt{7}x$ .

Using the law of sines on $\triangle{ABC} \implies \dfrac{3x}{\sin(\lambda)} = \dfrac{\sqrt{7}x}{\sin(60^{\circ})} = \dfrac{2\sqrt{7}x}{\sqrt{3}} \implies$

$\sin(\lambda) = \dfrac{3}{2}\sqrt{\dfrac{3}{7}}$

Using the law of sines on $\triangle{DAC} \implies \dfrac{50}{\sin(\lambda)} = \dfrac{n}{\sin(30^{\circ})} = 2n \implies$

$n = \dfrac{25}{\sin(\lambda)} = \boxed{\dfrac{50}{3}\sqrt{\dfrac{7}{3}}}$

Similarly using the law of sines on $\triangle{BAD} \implies m = \dfrac{25}{\sin(\alpha)} =$ $\dfrac{25}{\sin(120^{\circ} - \lambda)} =$

$\dfrac{50}{\sqrt{3}\cos(\lambda) + \sin(\lambda)} = \boxed{25\sqrt{\dfrac{7}{3}}}$ , where $\cos(\lambda) = \dfrac{1}{2\sqrt{7}}$

$\implies y = m + n = \boxed{\dfrac{125}{3}\sqrt{\dfrac{7}{3}}} = \sqrt{7}x \implies$ $x = \dfrac{125}{3\sqrt{3}} \implies 2x = \dfrac{250}{3\sqrt{3}} \implies$

$h = 2x\sin(\lambda) = \boxed{\dfrac{125}{\sqrt{7}}} \implies$ $A_{\triangle{ABC}} = \dfrac{1}{2}yh = \dfrac{125^2}{2 * 3\sqrt{3}} = \dfrac{5^{(2 * 3)}}{2 * 3\sqrt{3}} =$

$\dfrac{a^{c * b}}{c * b\sqrt{b}} \implies a + b + c = \boxed{10}$ .

Using the cosine rule on the three triangles:

$(3x)^2 + 50^2 - y^2 = 2\cdot3x\cdot50\cdot\cos \frac{\pi}{6}$

$(2x)^2 + 50^2 - z^2 = 2\cdot2x\cdot50\cdot\cos \frac{\pi}{6}$

$(3x)^2 + (2x)^2 - (y+z)^2 = 2\cdot3x\cdot2x\cdot\cos\frac{\pi}{3}$

Three equations three unkowns, solve for the sides and compute the area with heron's formula (see below):

$\triangle{ACD} = \dfrac{15625\sqrt{3}}{18} = \dfrac{5^{3\cdot2}}{3\cdot2\sqrt{3}} \implies a=5, b=3,c=2\implies a+b+c=\boxed{10}$

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from sympy import *
from sympy.abc import x, y, z

def heron(a, b, c):
    s = (a + b + c)/2 
    return sqrt(s*(s - a)*(s - b)*(s - c))

M = 50 
eq1 = (3*x)**2 + M**2 - y**2 - 2 * 3*x * M * cos(pi/6)
eq2 = (2*x)**2 + M**2 - z**2 - 2 * 2*x * M * cos(pi/6)
eq3 = (3*x)**2 + (2*x)**2 - (y + z)**2 - 2 * 3*x * 2*x * cos(pi/3)
sol = solve([eq1, eq2, eq3], [x,y,z])
for x, y, z in sol:
    area = heron(3*x, 2*x, y + z)
    if area > 0:
        print(simplify(area))

1

15625*sqrt(3)/18

blah $\dfrac{15625\sqrt{3}}{18} = \dfrac{5^{3\cdot2}}{3\cdot2\sqrt{3}}$

Similar solution as @Rocco Dalto's

By sine rule ,

$\begin{cases} \dfrac {AB}{AD} = \dfrac {\sin \angle ADB}{\sin B} = \dfrac {\sin (180^\circ - \angle ADB}{\sin B} = \dfrac {\sin \angle ADC}{\sin B} & ...(1) \\ \dfrac {AC}{AD} = \dfrac {\sin \angle ADC}{\sin C} & ...(2) \end{cases}$

From $\dfrac {(1)}{(2)}$ :

$\begin{aligned} \frac {AB}{AC} & = \frac {\sin \blue C}{\sin B} & \small \blue{\text{Note that }C = 180^\circ - 60^\circ - B} \\ \frac 32 & = \frac {\sin (120^\circ -B)}{\sin B} \\ 3 \sin B & = 2 \sin (120^\circ - B) = \sqrt 3 \cos B + \sin B \\ \implies \tan B & = \frac {\sqrt 3}2 \end{aligned}$

From $(1)$ :

$\begin{aligned} \frac {AB}{AD} & = \frac {\sin \blue{\angle ADC}}{\sin B} & \small \blue{\text{Note that }\angle ADC = B + 30^\circ} \\ \frac {AB}{50} & = \frac {\sin(B+30^\circ)}{\sin B} \\ & = \frac {\frac 12 \cos B + \frac {\sqrt 3}2 \sin B}{\sin B} \\ \frac {AB}{25} & = \cot B + \sqrt 3 = \frac 2{\sqrt 3} + \sqrt 3 \\ \implies AB & = \frac {5^3}{\sqrt 3} \end{aligned}$

The area of $\triangle ABC$ : $A_{\triangle ABC} = \dfrac {AB\cdot AC \sin 60^\circ}2 = \dfrac {\frac 23AB^2 \cdot \frac {\sqrt 3}2}2 = \left(\dfrac {5^3}{\sqrt 3}\right)^2\cdot \dfrac 1{2\sqrt 3} = \dfrac {5^6}{6\sqrt 3}$ . Therefore $a+b+c = 5+2+3 = \boxed{10}$