A vector perpendicular to the plane containing the points \(A(1,-1,2), B(2,0,-1), C(0,2,1)\) is

We know that a vector perpendicular to the plane containing the points \(\mathbf{A}, \mathbf{B}, \mathbf{C}\) is given by \(\mathbf{A} \times \mathbf{B}+\mathbf{B} \times \mathbf{C}+\mathbf{C} \times \mathbf{A}\)
We have, \(\mathbf{A}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{B}=2 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) and \(\mathbf{C}=0 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Now, \(\mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 2 \\ 2 & 0 & -1\end{array}\right|=\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
\(\begin{aligned} \mathbf{B} \times \mathbf{C}=&\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 0 & -1 \\ 0 & 2 & 1\end{array}\right|=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ \mathbf{C} \times \mathbf{A}=&\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 2 & 1 \\ 1 & -1 & 2\end{array}\right|=5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ \text { Thus, } \mathbf{A} \times \mathbf{B}+\mathbf{B} \times \mathbf{C}+\mathbf{C} \times \mathbf{B}=(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ & \quad+(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})+(5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\=& 8 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned}\)