Let the vectors → u _ 1 = ^ i + ^ j + a ^ k , → u _ 2 = ^ i + b ^ j...

Given: \overset{\to}{u}1 = \overset{^}{i} + \overset{^}{j} + a\overset{^}{k}, \overset{\to}{u}2 = \overset{^}{i} + b\overset{^}{j} + \overset{^}{k} and \overset{\to}{u}3 = c\overset{^}{i} + \overset{^}{j} + \overset{^}{k} are coplanar.

Also given that,\overset{\to}{v}1 = \left(a + b\right)\overset{^}{i} + c\overset{^}{j} + c\overset{^}{k}, \overset{\to}{v}2 = a\overset{^}{i} + \left(b + c\right)\overset{^}{j} + a\overset{^}{k}and \overset{\to}{v}3 = b\overset{^}{i} + b\overset{^}{j} + \left(c + a\right)\overset{^}{k} are coplanar.

Now, using the condition of coplanar we get,

⇒ \left|\begin{cases}1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1\end{cases}\right| = 0

Expanding the determinant along ${R}_{1}$.

$⇒ \left(b - 1\right) - \left(1 - c\right) + a\left(1 - bc\right) = 0$

$⇒ a + b + c = 2 + abc . . . . . \left(i\right)$

Again using the coplanar condition we get,

⇒ \left|\begin{cases}a + b & c & c \\ a & b + c & a \\ b & b & c + a\end{cases}\right| = 0

Apply row transformations, \left({R}_{3} \to {R}_{3} - \left({R}_{1} + {R}_{2}\right)\right)$</p><p>$⇒ \left|$\begin{cases}a + b & c & c \\ a & b + c & a \\ - 2a & - 2c & 0\end{cases}$\right| = 0$</p><p>Expand the determinant along${R}_{1}$.</p><p>$⇒ \left(a + b\right)\left(0 + 2ac\right) - c\left(0 + 2{a}^{2}\right) + c\left(- 2ac + 2a\left(b + c\right)\right) = 0$</p><p>$⇒ 2{a}^{2}c + 2abc - 2{a}^{2}c - 2a{c}^{2} + 2abc + 2a{c}^{2} = 0$</p><p>$⇒ abc = 0$</p><p>$∴ a + b + c = 2$(From eq \left(i\right)$)

$∴ 6\left(a + b + c\right) = 12$

Hence this is the correct option.