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Observe that
[tex]a^2 + b^2 + ab = 7 \iff a^2 + b^2 - 2\cos(120^\circ) = \left(\sqrt 7\right)^2[/tex]
[tex]b^2 + c^2 + bc = 21 \iff b^2 + c^2 - 2\cos(120^\circ) = \left(\sqrt{21}\right)^2[/tex]
[tex]a^2 + c^2 + ac = 28 \iff a^2 + c^2 - 2\cos(120^\circ) = \left(\sqrt{28}\right)^2[/tex]
Drawing a comparison to the law of cosines, we see the first equation defines a triangle with sides [tex]a[/tex], [tex]b[/tex], and [tex]\sqrt 7[/tex]; the second equation defines one with sides [tex]b[/tex], [tex]c[/tex], and [tex]\sqrt{21}[/tex]; and the third equation defines another one with sides [tex]a[/tex], [tex]c[/tex], and [tex]\sqrt{28}[/tex]. We can then join these triangles at a point D to form a new triangle ABC with sides [tex]\sqrt7,\sqrt{21},\sqrt{28}[/tex]. (See attached sketch)
Now, the area of ABC is the sum of the areas of ABD, BCD, and ACD. The area of any of these triangles is equal to half the area of a parallelogram spanned by the sides involving the vertex D. The area of a parallelogram with adjacent sides [tex]x[/tex] and [tex]y[/tex] and with angle [tex]\theta[/tex] between them is [tex]ab\sin(\theta)[/tex].
For example, triangle ABD has area
[tex]\mathrm{area}_{\Delta ABD} = \dfrac12 \times ab \sin(120^\circ)[/tex]
Then the total area of triangle ABC is
[tex]\mathrm{area}_{\Delta ABC} = \dfrac12 \sin(120^\circ) (ab + ac + bc)[/tex]
We use Heron's formula to find the area of ABC. If [tex]s[/tex] is its semiperimeter, i.e.
[tex]s = \dfrac{\sqrt7 +\sqrt{21} + \sqrt{28}}2[/tex]
then, with a lot of algebraic simplification,
[tex]\mathrm{area}_{\Delta ABC} = \sqrt{s (s-\sqrt7) (s - \sqrt{21}) (s - \sqrt{28})} = \dfrac{7\sqrt3}2[/tex]
Finally, we find
[tex]\dfrac{7\sqrt3}2 = \dfrac12 \sin(120^\circ) (ab + ac + bc) \implies ab + ac + bc = \dfrac{\frac{7\sqrt3}2}{\frac12\times\frac{\sqrt3}2} = \boxed{14}[/tex]
