Explanation
\begin{aligned}
& \mathrm{Since,~}\vec{u}_{1},\vec{u}_{2},\vec{u}_{3}\text{ are coplanar.} \\
& \mathrm{So},[\vec{u}_{1}\vec{u}_{2}\vec{u}_{3}]=0 \\
& \Rightarrow
\begin{vmatrix}
1 & 1 & a \\
1 & b & 1 \\
c & 1 & 1
\end{vmatrix}=0 \\
& \Rightarrow1(b-1)-1(1-c)+a(1-bc)=0 \\
& \Rightarrow a+b+c-2=abc...(\mathbf{i}) \\
& \mathrm{Also,}\left[\vec{v}_{1}\vec{v}_{2}\vec{v}_{3}\right]=0 \\
& \Rightarrow
\begin{vmatrix}
a+b & c & c \\
a & b+c & a \\
b & b & c+a
\end{vmatrix}=0 \\
& \Rightarrow(a+b)\left[bc+ba+c^{2}+ca-ab\right]-c\left[ac+a^{2}-ab\right] \\
& +c[ab-b^{2}-bc]=0 \\
& \Rightarrow abc+ac^{2}+a^{2}c+b^{2}c+bc^{2}+abc-ac^{2}-a^{2}c \\
& +abc+abc-b^{2}c-bc^{2}=0 \\
& \Rightarrow4abc=0\Rightarrow abc=0...(\mathrm{ii}) \\
& \mathrm{So},a+b+c-2=0[\mathrm{from~(i)}] \\
& \Rightarrow a+b+c=2 \\
& \Rightarrow6(a+b+c)=12
\end{aligned}
