Concept:

\(\text { Let } \overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \overrightarrow{\mathrm{i}}+\mathrm{b}_{1} \overrightarrow{\mathrm{j}}+\mathrm{c}_{1} \overrightarrow{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\mathrm{a}_{2} \overrightarrow{\mathrm{i}}+\mathrm{b}_{2} \overrightarrow{\mathrm{j}}+\mathrm{c}_{2} \overrightarrow{\mathrm{k}} \text { and } \overrightarrow{\mathrm{c}}=\mathrm{a}_{3} \overrightarrow{\mathrm{i}}+\mathrm{b}_{3} \overrightarrow{\mathrm{j}}+\mathrm{c}_{3} \overrightarrow{\mathrm{k}} \text { be the three vectors }\)

Condition for coplanarity:

\( \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\left|\begin{array}{lll} \rm a_{1} & \mathrm{b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{b}_{3} & \mathrm{c}_{3} \end{array}\right|=0 \)

​Calculation:

Here, \(\rm \hat i-\hat j+\hat k, 2\hat i+\hat j-\hat k, \hat iλ-\hat j+\hat k λ \) are coplanar

\(\begin{array}{l} \Rightarrow \left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ λ & -1 & λ \end{array}\right|=0\end{array}\)

1(λ - 1) + 1(2λ + λ) + 1(-2 - λ) = 0

λ - 1 + 2λ + λ + -2 - λ = 0

3λ - 3 = 0

λ = 0

Hence, option (4) is correct.