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If $\quad \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \quad \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}} \quad$ and $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=\lambda \overrightarrow{\mathbf{a}}+\mu \overrightarrow{\mathbf{b}}$, then $\lambda+\mu$ is equal to:
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We have
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|$
$=\hat{\mathbf{i}}(-1)-\hat{\mathbf{j}}(-1)+\hat{\mathbf{k}}(1-1)$
$=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$
$(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=\left|\begin{array}{rrr}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right|=\hat{\mathbf{k}}(-1)=-\hat{\mathbf{k}}$
Now, $\lambda \overrightarrow{\mathbf{a}}+\mu \overrightarrow{\mathbf{b}}=\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+\hat{\mathbf{k}})$
$=(\lambda+\mu) \hat{\mathbf{i}}+(\lambda+\mu) \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$
$\because \quad \lambda \overrightarrow{\mathbf{a}}+\mu \overrightarrow{\mathbf{b}}=(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}$
$\Rightarrow(\lambda+\mu) \hat{\mathbf{i}}+(\lambda+\mu) \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}=-\hat{\mathbf{k}}$
Equating the coefficient of $\hat{\mathbf{i}}$
$\therefore \quad \lambda+\mu=0$
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|$
$=\hat{\mathbf{i}}(-1)-\hat{\mathbf{j}}(-1)+\hat{\mathbf{k}}(1-1)$
$=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$
$(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=\left|\begin{array}{rrr}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right|=\hat{\mathbf{k}}(-1)=-\hat{\mathbf{k}}$
Now, $\lambda \overrightarrow{\mathbf{a}}+\mu \overrightarrow{\mathbf{b}}=\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+\hat{\mathbf{k}})$
$=(\lambda+\mu) \hat{\mathbf{i}}+(\lambda+\mu) \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$
$\because \quad \lambda \overrightarrow{\mathbf{a}}+\mu \overrightarrow{\mathbf{b}}=(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}$
$\Rightarrow(\lambda+\mu) \hat{\mathbf{i}}+(\lambda+\mu) \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}=-\hat{\mathbf{k}}$
Equating the coefficient of $\hat{\mathbf{i}}$
$\therefore \quad \lambda+\mu=0$
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