If \(\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\) and \(\overline{\mathrm{c}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) are such that \((\bar{a}+\lambda \bar{b})\) is perpendicular to \(\bar{c}\), then the value of \(\lambda\) is

Let \(\overline{\mathrm{d}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\)
\(\begin{aligned}
\overline{\mathrm{d}} & =(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\
& =2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+2 \lambda \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}} \\
& =(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}
\end{aligned}\)
Now, \(\overline{\mathrm{d}}\) is perpendicular to \(\overline{\mathrm{c}}\).
\(\begin{aligned}
& \therefore \quad \overline{\mathrm{c}} \cdot \overline{\mathrm{d}}=0 \\
& \Rightarrow(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot[(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}]=0 \\
& \Rightarrow 1(2 \lambda+2)+3(3+\lambda)=0 \\
& \Rightarrow 2 \lambda+2+9+3 \lambda=0 \\
& \Rightarrow 5 \lambda+11=0 \\
& \Rightarrow \lambda=\frac{-11}{5}
\end{aligned}\)