Let the vectors $$(2 + a + b)\hat{i} + (a + 2b+c)\hat{j} - (b + c)\hat{k}$$, $$(1 + b)\hat{i}+2b\hat{j}-b\hat{k}$$ and $$(2 + b)\hat{i} + 2b\hat{j} + (1 - b)\hat{k}$$, $$ a, b, c \in R$$ be co-planar. Then which of the following is true?

Given vectors

$$\vec v_1=(2+a+b)\hat i+(a+2b+c)\hat j-(b+c)\hat k$$

$$\vec v_2=(1+b)\hat i+2b\hat j-b\hat k$$

and

$$\vec v_3=(2+b)\hat i+2b\hat j+(1-b)\hat k$$

Since the vectors are coplanar,

$$\begin{vmatrix}2+a+b & a+2b+c & -(b+c)\\1+b & 2b & -b\\2+b & 2b & 1-b\end{vmatrix}=0$$

Apply

$$R_3\to R_3-R_2$$

Then determinant becomes

$$\begin{vmatrix}2+a+b & a+2b+c & -(b+c)\\1+b & 2b & -b\\1 & 0 & 1\end{vmatrix}=0$$

Expanding along third row,

Cofactor expansion formula is

$$\sum a_{ij}(-1)^{i+j}M_{ij}$$

For third row:

- element at position

$$(3,1)=1$$

Its sign is

$$(-1)^{3+1}=+1$$

Minor obtained after deleting third row and first column:

$$\begin{vmatrix} a+2b+c & -(b+c)\\ 2b & -b \end{vmatrix}$$

So first term is

$$+1\cdot\begin{vmatrix}a+2b+c & -(b+c)\\2b & -b\end{vmatrix}$$

Now second element of third row is

$$(3,2)=0$$

so its contribution is zero.

Third element is

$$(3,3)=1$$

Its sign is

$$(-1)^{3+3}=+1$$

Minor obtained after deleting third row and third column:

$$\begin{vmatrix} 2+a+b & a+2b+c\\1+b & 2b\end{vmatrix}$$

Hence third term is

$$+1\cdot\begin{vmatrix}2+a+b & a+2b+c\\1+b & 2b\end{vmatrix}$$

$$1\cdot\begin{vmatrix}a+2b+c & -(b+c)\\2b & -b\end{vmatrix}+1\cdot\begin{vmatrix}2+a+b & a+2b+c\\1+b & 2b\end{vmatrix}=0$$

Now,

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

$$=-ab-2b^2-bc+2b^2+2bc$$

$$=bc-ab$$

Also,

$$\begin{vmatrix}2+a+b & a+2b+c\\1+b & 2b\end{vmatrix}=(2+a+b)(2b)-(1+b)(a+2b+c)$$

$$=4b+2ab+2b^2-a-2b-c-ab-2b^2-bc$$

$$=2b+ab-a-c-bc$$

Adding,

$$bc-ab+2b+ab-a-c-bc=0$$

$$2b-a-c=0$$

Hence,

$$a+c=2b$$