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 $$6(a + b + c)$$ is equal to

Vectors are coplanar if their scalar triple product (determinant) is zero:

$$\begin{vmatrix} 1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1 \end{vmatrix} = 0$$

Expanding along the first row:

$$1(b - 1) - 1(1 - c) + a(1 - bc) = 0$$

$$b - 1 - 1 + c + a - abc = 0 \implies \mathbf{a + b + c - 2 = abc}$$

$$\begin{vmatrix} a+b & c & c \\ a & b+c & a \\ b & b & c+a \end{vmatrix} = 0$$

Using row operations ($$R_1 \to R_1 - R_2 - R_3$$):

$$\begin{vmatrix} 0 & -2b & -2a \\ a & b+c & a \\ b & b & c+a \end{vmatrix} = 0$$

Expanding along the first row:

$$2b(a(c+a) - ab) - 2a(ab - b(b+c)) = 0$$

$$2b(ac + a^2 - ab) - 2a(ab - b^2 - bc) = 0$$

Divide by $$2b$$ (assuming $$b \neq 0$$):

$$(ac + a^2 - ab) - \frac{a}{b}(ab - b^2 - bc) = 0$$

$$ac + a^2 - ab - a^2 + ab + ac = 0 \implies \mathbf{2ac = 0}$$

This implies $$a=0$$ or $$c=0$$.

If $$a=0$$, substituting into our first equation ($$a + b + c - 2 = abc$$):

$$0 + b + c - 2 = 0 \implies \mathbf{b + c = 2}$$

The expression we need is $$6(a + b + c)$$:

$$6(0 + 2) = \mathbf{12}$$