The scalar product of the vector \(\hat{i}+\hat{j}+\hat{k}\) with a unit vector along the sum of the vectors \(2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\lambda \hat{i}+2 \hat{j}+3 \hat{k}\) is equal to 1 , then value of \(\lambda\) is

\((2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})+(\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=(2+\lambda) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
Unit vector along the above vector is
\(\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}\)
Scalar product of \((\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\) with this unit vector is 1 .
\(\begin{aligned}
\therefore \quad & (\hat{i}+\hat{j}+\hat{k}) \cdot \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}=1 \\
& \frac{(2+\lambda)+6-2}{\sqrt{\lambda^2+4 \lambda+44}}=1 \\
& \sqrt{\lambda^2+4 \lambda+44}=\lambda+6 \\
\therefore \quad & \lambda^2+4 \lambda+44=(\lambda+6)^2 \\
& 8 \lambda=8 \\
& \lambda=1
\end{aligned}\)