Let u_1= i+ j+a k, u_2= i+b j+ k, u_3=c i+ j+ k be coplanar.

Question

Let $\vec u_1=\hat i+\hat j+a\hat k$, $\vec u_2=\hat i+b\hat j+\hat k$, $\vec u_3=c\hat i+\hat j+\hat k$ be coplanar. If $\vec v_1=(a+b)\hat i+c\hat j+c\hat k$, $\vec v_2=a\hat i+(b+c)\hat j+a\hat k$, $\vec v_3=b\hat i+b\hat j+(c+a)\hat k$ are also coplanar, then $6(a+b+c)$ is:

Accepted Answer

From $[\vec u_1\ \vec u_2\ \vec u_3]=0$: $\begin{vmatrix}1&1&a\1&b&1\c&1&1\end{vmatrix}=0 \implies (b-1)+(c-1)+a(1-bc)=0 \implies abc = a+b+c-2$.

From $[\vec v_1\ \vec v_2\ \vec v_3]=0$: expanding $\begin{vmatrix}a+b&c&c\a&b+c&a\b&b&c+a\end{vmatrix}=0$ simplifies to $4abc=0 \Rightarrow abc=0$.

Combining: $a+b+c-2=0 \Rightarrow a+b+c=2$, so $6(a+b+c)=12$.

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