Let the vectors $\vec{u}_{1}=\hat{i}+\hat{j}+a \hat{k}, \vec{u}_{2}=\hat{i}+b \hat{j}+\hat{k}$ and $\vec{u}_{3}=c \hat{i}+\hat{j}+\hat{k}$ be coplanar. If the vectors $$\vec{v}_{1}=(a+b) \hat{i}+c \hat{j}+c \hat{k}, \vec{v}_{2}=a \hat{i}+(b+c) \hat{j}+a \hat{k}$$ and $\vec{v}_{3}=b \hat{i}+b \hat{j}+(c+a) \hat{k}$ are also coplanar, then $6(\mathrm{a}+\mathrm{b}+\mathrm{c})$ is equal to :

Since, $\vec{u}_1, \vec{u}_2, \vec{u}_3$ are coplanar. <br/><br/>So, $\left[\begin{array}{lll}\vec{u}_1 & \vec{u}_2 & \vec{u}_3\end{array}\right]=0$ <br/><br/>$$ \begin{aligned} & \Rightarrow\left|\begin{array}{lll} 1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1 \end{array}\right|=0 \\\\ & \Rightarrow 1(b-1)-1(1-c)+a(1-b c)=0 \\\\ & \Rightarrow b-1-1+c+a-a b c=0 \\\\ & \Rightarrow a+b+c-2=a b c ........... (i) \end{aligned} $$ <br/><br/>Also, $\left[\begin{array}{lll}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{array}\right]=0$ <br/><br/>$$ \begin{aligned} & \Rightarrow\left|\begin{array}{ccc} a+b & c & c \\ a & b+c & a \\ b & b & c+a \end{array}\right|=0 \\ & \Rightarrow(a+b)\left[b c+b a+c^2+c a-a b\right]-c\left[a c+a^2-a b\right] \\\\ & \quad+c\left[a b-b^2-b c\right]=0 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow a b c+a c^2+a^2 c+b^2 c+b c^2+a b c-a c^2-a^2 c \\\\ & +a b c+a b c-b^2 c-b c^2=0 \\\\ & \Rightarrow 4 a b c=0 \Rightarrow a b c=0 \\\\ & \begin{array}{lr} \text { So, } a+b+c-2=0 [from (i)]\\\\ \Rightarrow a+b+c=2 \end{array} \\\\ & \Rightarrow 6(a+b+c)=12 \end{aligned} $$