The scalar product of vectors \(\bar{a}=\hat{i}+2 \hat{j}+\hat{k}\) and a unit vector along the sum of vectors \(\overline{\mathrm{b}}=2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) and \(\overline{\mathrm{c}}=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\) is one, then the value of \(\lambda\) is

\(\overline{\mathrm{b}}+\overline{\mathrm{c}}=(2+\lambda) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \)
\( \text {Unit vector }=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{\overline{\mathrm{b}}+\overline{\mathrm{c}}}} \)
\( =\frac{(2+\lambda) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{(2+\lambda)^2+(-2)^2+2^2}} \)
\( =\frac{(2+\lambda) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{\lambda^2+4 \lambda+12}}\)
According to the given condition,
\((\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot\left(\frac{(2+\lambda) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{\lambda^2+4 \lambda+12}}\right)=1 \)
\( \Rightarrow \frac{(2+\lambda)-4+2}{\sqrt{\lambda^2+4 \lambda+12}}=1 \)
\( \Rightarrow \lambda=\sqrt{\lambda^2+4 \lambda+12} \)
\( \Rightarrow \lambda^2=\lambda^2+4 \lambda+12 \)
\( \Rightarrow 4 \lambda=-12 \)
\( \Rightarrow \lambda=-3\)