Let the vectors \(\mathbf{u}_1 = \hat{i} + \hat{j} + a\hat{k}, \mathbf{u}_2 = \hat{i} + b\hat{j} + \hat{k}\), and \(\mathbf{u}_3 = c\hat{i} + \hat{j} + \hat{k}\) be coplanar. If the vectors \(\mathbf{v}_1 = (a + b)\hat{i} + c\hat{j} + c\hat{k}, \mathbf{v}_2 = a\hat{i} + (b + c)\hat{j} + a\hat{k}, \mathbf{v}_3 = b\hat{i} + b\hat{j} + (c + a)\hat{k}\) are also coplanar, then \(6(a + b + c)\) is equal to:
Show Hint
The scalar triple product of three vectors being zero implies that the vectors are coplanar. Utilize this property to solve problems involving coplanarity.
For the vectors \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \) to be coplanar:
\[
[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3] =
\begin{vmatrix}
a + b & c & c \\
a & b + c & a \\
b & b & c + a
\end{vmatrix} = 0.
\]
Perform row operations:
\[
R_3 \to R_3 - (R_1 + R_2).
\]
The matrix becomes:
\[
\begin{vmatrix}
a + b & c & c \\
a & b + c & a \\
-a & b + c - 2a & -2c
\end{vmatrix}.
\]
Expanding along the first row:
\[
[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3] = (a + b)
\begin{vmatrix}
b + c & a \\
b + c - 2a & -2c
\end{vmatrix} - c
\begin{vmatrix}
a & a \\
b & -2c
\end{vmatrix} + c
\begin{vmatrix}
a & b + c \\
b & b + c - 2a
\end{vmatrix}.
\]
