If the area of the triangle with vertices \(\hat{i}+y \hat{j}, \hat{i}+2 \hat{k}\) and \(3 \hat{j}+\hat{k}\) is \(\sqrt{6}\) sq. units, then the values of \(y\) are

\(\begin{aligned} & \overrightarrow{A B}=(\hat{i}+2 \hat{k})-(\hat{i}+y \hat{j})=0 \hat{i}-y \hat{j}+2 \widehat{k} \\ & \overrightarrow{B C}=(3 \hat{j}+\hat{k})-(\hat{i}+2 \hat{k})=-\hat{i}+3 \hat{j}-\hat{k} \\ & \text { Now area of triangle }=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{6} \\ & \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 0 & -y & 2 \\ -1 & 3 & -1\end{array}\right|=\sqrt{6} \\ & \Rightarrow|(y-6) \hat{i}-2 \hat{j}-y \hat{k}|=2 \sqrt{6} \\ & \Rightarrow(y-6)^2+(-2)^2+(-y)^2=(2 \sqrt{6})^2 \\ & \Rightarrow y^2-12 y+36+4+y^2=24 \\ & \Rightarrow y^2-6 y+8=0 \\ & \Rightarrow y=2,4\end{aligned}\)