The vector(s) which is/are coplanar with vectors \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\), are perpendicular to the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) is/are

Let \(\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
\(\therefore\) A vector coplanar to \(\mathbf{a}\) and \(\mathbf{b}\), and perpendicular to \(\mathbf{c}\).
Now, \(\quad \lambda(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\)
\(\Rightarrow \quad \lambda\{(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}\}\)
\(\Rightarrow \quad \lambda\{(1+1+4)(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\)
\(-(1+2+1)(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})\}\)
\(\Rightarrow \lambda\{6 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}-6 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}\}\)
\(\Rightarrow \lambda\{6 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\} \Rightarrow 6 \lambda(\hat{\mathbf{j}}-\hat{\mathbf{k}})\)
For \(\lambda=\frac{1}{6} \Rightarrow\) Option (a) is correct.
For \(\lambda=-\frac{1}{6} \Rightarrow\) Option (d) is correct.