Add all the displacement vectors together, then find the magnitude of this vector sum.
[tex]\vec r_1=(72\,\mathrm{km})(\cos60^\circ\,\vec\imath+\sin60^\circ\,\vec\jmath)[/tex]
[tex]\vec r_2=(48\,\mathrm{km})(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)[/tex]
[tex]\vec r_3=(100\,\mathrm{km})(\cos150^\circ\,\vec\imath+\sin150^\circ\,vec\jmath)[/tex]
Summing these vectors gives the resultant displacement
[tex]\vec r=\vec r_1+\vec r_2+\vec r_3\approx(-50.6\,\vec\imath+64.4\,\vec\jmath)\,\mathrm{km}[/tex]
and we have
[tex]\|\vec r\|=\sqrt{(-50.6)^2+64.4^2}\,\mathrm{km}\approx\boxed{81.9\,\mathrm{km}}[/tex]
so the plane ends up about 81.9 km away from its starting position.