Let the vectors vec{u} = (2+a+b) hat{i}+(a+2 | vedclass

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(B) For the vectors to be co-planar,their scalar triple product must be zero:$\begin{vmatrix} a+b+2 & a+2 b+c & -(b+c) \ b+1 & 2 b & -b \ b+2 & 2 b

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$2 b=a+c$

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(B) For the vectors to be co-planar,their scalar triple product must be zero:$\begin{vmatrix} a+b+2 & a+2 b+c & -(b+c) \ b+1 & 2 b & -b \ b+2 & 2 b & 1-b \end{vmatrix} = 0$Applying row operations $R_3 ightarrow R_3 - R_2$ and $R_1 ightarrow R_1 - R_2$:$\begin{vmatrix} a+1 & a+c & -c \ b+1 & 2 b & -b \ 1 & 0 & 1 \end{vmatrix} = 0$Expanding along the third row:$1((a+c)(-b) - (2b)(-c)) + 1((a+1)(2b) - (b+1)(a+c)) = 0$$(-ab - bc + 2bc) + (2ab + 2b - (ab + ac + b + c)) = 0$$(-ab + bc) + (ab - ac + b - c) = 0$$bc - ac + b - c = 0$$c(b-a) + 1(b-c) = 0$Wait,let us re-evaluate the determinant expansion:$(a+1)(2b) - (a+c)(b+1+b) + (-c)(-2b) = 0$$(a+1)(2b) - (a+c)(2b+1) + 2bc = 0$$2ab + 2b - (2ab + a + 2bc + c) + 2bc = 0$$2ab + 2b - 2ab - a - 2bc - c + 2bc = 0$$2b - a - c = 0$Therefore,$2b = a+c$.

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

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