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If $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathbf{j}}$ such that $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda$.
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$\begin{array}{ll}\overrightarrow{\mathrm{a}}= & 2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \therefore \quad & \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & =(2-\lambda) \hat{\mathrm{i}}+(2+2 \lambda) \hat{\mathrm{j}}+(3+\lambda) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}, \quad \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}} \text { is pendicular to } \overrightarrow{\mathrm{c}} \\ \Rightarrow \quad & (\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=0 \\ & 3(2-\lambda)+(2+2 \lambda)=0 \\ \Rightarrow \quad & 6-3 \lambda+2+2 \lambda=0 \text { or } \lambda=8\end{array}$
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