$$\begin{gathered} Consider three vectors \text{a}=\widehat{i}+\widehat{j}+\widehat{k},\text{\hspace{0.17em}b}=\widehat{i}+\widehat{j}-\widehat{k}\text{\hspace{0.17em}and\hspace{0.17em}c}=\widehat{i}-\widehat{j}.If \\ {\text{k}}_{\text{1}}\text{a}+{\text{k}}_{\text{2}}\text{b}+{\text{k}}_{\text{3}}\text{c}=4\widehat{i}+6\widehat{j}+\widehat{k} where{\text{k}}_{\text{1}}{\text{,k}}_{\text{2\hspace{0.17em}}}{\text{and\hspace{0.17em}k}}_{\text{3}}\text{ are scalars. Then} \end{gathered}$$

$$\begin{gathered} {\text{k}}_{\text{1}}\text{a}+{\text{k}}_{\text{2}}\text{b}+{\text{k}}_{\text{3}}\text{c}=4\widehat{i}+6\widehat{j}+\widehat{k} \\ {\text{k}}_{\text{1}}\left(\widehat{i}+\widehat{j}+\widehat{k}\right)+{\text{k}}_{\text{2}}\left(\widehat{i}+\widehat{j}-\widehat{k}\right)+{\text{k}}_{\text{3}}\left(\widehat{i}-\widehat{j}\right)=4\widehat{i}+6\widehat{j}+\widehat{k} \\ {\text{k}}_{\text{1}}+{\text{k}}_{\text{2}}+{\text{k}}_{\text{3}}=4\to \left(1\right) \\ {\text{k}}_{\text{1}}+{\text{k}}_{\text{2}}-{\text{k}}_{\text{3}}=6\to \left(2\right) \\ {k}_1-k_2=1\to \left(3\right) \\ {k}_1+k_2=5\to \left(4\right) \\ From \left(1\right) and \left(2\right) \\ After solving ,we get \\ {\text{k}}_{\text{1}}=3 \\ {\text{k}}_{\text{1}}-{\text{k}}_{\text{2}}=1;{\text{\hspace{0.17em}k}}_{\text{2}}=2 \\ {\text{k}}_{\text{1}}+{\text{k}}_{\text{2}}-{\text{k}}_{\text{3}}=6\Rightarrow 3+2-{\text{k}}_{\text{3}}=6 \\ {\text{k}}_{\text{3}}=-1 \end{gathered}$$