The values of 'a' for which the vectors $\vec{\alpha }= \hat{i}+2\hat{j} + \hat{k}, \,\, \vec{\beta}=a\hat{i} + \hat{j}+2\hat{k}$ and $\vec{\gamma }=\hat{i}+2\hat{j}+a\hat{k}$ are coplanar are :

$1, \frac{1}{2}$

The correct answer is Option (3) → $1, \frac{1}{2}$

finding 'a' to satisfy coplanarity

$Δ=\begin{vmatrix}1&2&1\\a&1&2\\1&2&a\end{vmatrix}=0$ (for coplanarity to exist)

$R_1→R_1-2R_2\\R_3→R_3-2R_2$

$Δ=\begin{vmatrix}1-2a&0&-3\\a&1&2\\1-2a&0&a-4\end{vmatrix}$

$⇒(a-4)(1-2a)+3(1-2a)=0$

$a-2a^2-4+8a+3-6a=0$

$-2a^2-1+3a=0$

$2a^2-3a+1=0$

$2a^2-2a-a+1=0$

$2a(a-1)-1(a-1)=0$

$a=1,\frac{1}{2}$