If $\overrightarrow{\mathrm{a}}$ satisfies  $\overrightarrow{\mathrm{a}}\times (\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\widehat{\mathrm{k}})=\widehat{\mathrm{i}}-\widehat{\mathrm{k}},\text{ then }\overrightarrow{\mathrm{a}}$ is equal to

• A

  $\mathrm{\lambda }\widehat{\mathrm{i}}+(2\mathrm{\lambda }-1)\widehat{\mathrm{j}}+\mathrm{\lambda }\widehat{\mathrm{k}},\mathrm{\lambda }\in \mathrm{R}$

• B

  $\mathrm{\lambda }\widehat{\mathrm{i}}+(1-2\mathrm{\lambda })\widehat{\mathrm{j}}+\mathrm{\lambda }\widehat{\mathrm{k}},\mathrm{\lambda }\in \mathrm{R}$

• C

  $\mathrm{\lambda }\widehat{\mathrm{i}}+(2\mathrm{\lambda }+1)\widehat{\mathrm{j}}+\mathrm{\lambda }\widehat{\mathrm{k}},\mathrm{\lambda }\in \mathrm{R}$

• D

  $\mathrm{\lambda }\widehat{\mathrm{i}}-(1+2\mathrm{\lambda })\widehat{\mathrm{j}}+\mathrm{\lambda }\widehat{\mathrm{k}},\mathrm{\lambda }\in \mathrm{R}$