Let the vectors u_1= i+ j+a k, u_2= i+b j+ k, and u_3=c i+ j+ k be coplanar. If the vectors v_1=(a+b) i+c j+c k, v_2=a i+(b+c) j+a k and v_3=b i+b…

Given: ${\vec{u}}_{1}=\hat{i}+\hat{j}+a\hat{k},{\vec{u}}_{2}=\hat{i}+b\hat{j}+\hat{k}$u1​=i^+j^​+ak^,u2​=i^+bj^​+k^ and ${\vec{u}}_{3}=c\hat{i}+\hat{j}+\hat{k}$u3​=ci^+j^​+k^ are coplanar.

Also given that,${\vec{v}}_{1}=(a+b)\hat{i}+c\hat{j}+c\hat{k},{\vec{v}}_{2}=a\hat{i}+(b+c)\hat{j}+a\hat{k}$v1​=(a+b)i^+cj^​+ck^,v2​=ai^+(b+c)j^​+ak^and ${\vec{v}}_{3}=b\hat{i}+b\hat{j}+(c+a)\hat{k}$v3​=bi^+bj^​+(c+a)k^ are coplanar.

Now, using the condition of coplanar we get,

$$\Rightarrow |\begin{matrix}1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1\end{matrix}|=0$$⇒∣11c​1b1​a11​∣=0

Expanding the determinant along ${R}_{1}$R1​.

$$\Rightarrow (b-1)-(1-c)+a(1-bc)=0$$⇒(b−1)−(1−c)+a(1−bc)=0

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

Again using the coplanar condition we get,

$$\Rightarrow |\begin{matrix}a+b & c & c \\ a & b+c & a \\ b & b & c+a\end{matrix}|=0$$⇒∣a+bab​cb+cb​cac+a​∣=0

Apply row transformations, $({R}_{3}\rightarrow {R}_{3}-({R}_{1}+{R}_{2}))$(R3​→R3​−(R1​+R2​))

$$\Rightarrow |\begin{matrix}a+b & c & c \\ a & b+c & a \\ -2a & -2c & 0\end{matrix}|=0$$⇒∣a+ba−2a​cb+c−2c​ca0​∣=0

Expand the determinant along ${R}_{1}$R1​.

$$\Rightarrow (a+b)(0+2ac)-c(0+2{a}^{2})+c(-2ac+2a(b+c))=0$$⇒(a+b)(0+2ac)−c(0+2a2)+c(−2ac+2a(b+c))=0

$$\Rightarrow 2{a}^{2}c+2abc-2{a}^{2}c-2a{c}^{2}+2abc+2a{c}^{2}=0$$⇒2a2c+2abc−2a2c−2ac2+2abc+2ac2=0

$$\Rightarrow abc=0$$⇒abc=0

$\therefore a+b+c=2$∴a+b+c=2 (From eq $(i)$(i))

$$\therefore 6(a+b+c)=12$$∴6(a+b+c)=12

Hence this is the correct option.