Question about determine a point lies on a plane

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I was solving the following exercises.

Let $A,B,C$ be $3$ non-collinear points on a plane. Then, a point $M$ lies on the plane if and only if, there exists $a_1,a_2,a_3 \in R $ such that $\vec{OM} =a_1 \vec{OA} +a_2 \vec{OB} +a_3 \vec{OC} $, and $a_1+a_2+a_3=1$ , $O$ is an arbitrary point.

I came up with $2$ solutions.

No.1

Since $\vec{AM} , \vec{AB}, \vec{AC}$ coplanar,there must exists $b_1,b_2$ such that

$\vec{AM} = b_1*\vec{AB}+ b_2*\vec{AC}$

$\Rightarrow \vec{OM} -\vec{OA} =b_1*(\vec{OB}-\vec{OA})+ b_2*(\vec{OC}-\vec{OA})$

$\Rightarrow \vec{OM} =(1-b_1-b_2)*\vec{OA} + b_1*\vec{OB}+ b_2*\vec{OC}$

However,when I try another vector,

No.2

Since $\vec{OM} , \vec{AB}, \vec{AC}$ coplanar,there must exists $b_1,b_2$ such that

$\vec{OM} = b_1*\vec{AB}+ b_2*\vec{AC}$

$\Rightarrow \vec{OM} =b_1*(\vec{OB}-\vec{OA})+ b_2*(\vec{OC}-\vec{OA})$

$\Rightarrow \vec{OM} =(-b_1-b_2)*\vec{OA} + b_1*\vec{OB}+ b_2*\vec{OC}$

So, what wrong with the second solution ? Any help would be greatly appreciated! Thank you!