If A, B, C are the vertices of a triangle and X is a point in the plane of the triangle, then you can write: [MATH]\boldsymbol{x}=λ\boldsymbol{a}+μ\boldsymbol{b}+\nu \boldsymbol{c},\hspace1ex \text{ where }\lambda,\; \mu,\; \nu \; \text{ are scalars: } \lambda + \mu+\nu=1 \text{ and }\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}, \boldsymbol{x} \text{ are the position vectors of the points A, B, C, X respectively}.[/MATH]
Do you know that any vector in the plane can be written uniquely as a linear combination of 2 non-parallel vectors in the plane?
If so, assuming A, B, C form a (non-degenerate) triangle, then [MATH]\vec{CA}=\boldsymbol{a}-\boldsymbol{c}[/MATH], [MATH]\vec{CB}=\boldsymbol{b}-\boldsymbol{c}[/MATH] are two non-parallel vectors in the plane.
[MATH]\vec{CX}=\boldsymbol{x}-\boldsymbol{c}[/MATH], being a vector in the plane, can be written uniquely as [MATH]\hspace2ex λ(\boldsymbol{a}-\boldsymbol{c})+μ(\boldsymbol{b}-\boldsymbol{c})[/MATH]i.e. [MATH]\hspace1ex \exists! \hspace2ex \lambda \text{, } \mu: \hspace2ex \boldsymbol{x}-\boldsymbol{c}= λ(\boldsymbol{a}-\boldsymbol{c})+μ(\boldsymbol{b}-\boldsymbol{c})[/MATH]Now re-arrange that to get the form you are seeking and you should be ok.
(I hope I'm not begging the question!)