If \(\overline{\mathrm{a}}=\mathrm{m} \overline{\mathrm{b}}+\mathrm{n} \overline{\mathrm{c}}\), where \(\overline{\mathrm{a}}=4 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\), \(\overline{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\), then \(\mathrm{m}+\mathrm{n}=\)

Given:
\(\begin{aligned}
& \overline{\mathrm{a}}=4 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-18 \hat{\mathrm{k}} \\
& \overline{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
& \overline{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}
\end{aligned}\)
Also, \(\bar{a}=m \bar{b}+n \bar{c}\)
\(\begin{aligned}
& \Rightarrow 4 \hat{i}+13 \hat{j}-18 \hat{k}=m(\hat{i}-2 \hat{j}+3 \hat{k})+n(2 \hat{i}+3 \hat{j}-4 \hat{k}) \\
& \Rightarrow 4 \hat{i}+13 \hat{j}-18 \hat{k} \\
& =(m+2 n) \hat{i}+(-2 m+3 n) \hat{j}+(3 m-4 n) \hat{k}
\end{aligned}\)
Comparing, we get
\(m+2 n=4 \text { and }-2 m+3 n=13\)
Solving above equations, we get
\(\begin{aligned}
& \mathrm{m}=-2 \text { and } \mathrm{n}=3 \\
\therefore \quad & \mathrm{m}+\mathrm{n}=-2+3=1
\end{aligned}\)