The scalar product of the vector

$(2 \hat{i}+4 \hat{j}-5 \hat{k})+(\lambda \hat{i}+2 \hat{j}+3 \hat{k})$

$=(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}$

Therefore, unit vector along $(2 \hat{i}+4 \hat{j}-5 \hat{k})+(\lambda \hat{i}+2 \hat{j}+3 \hat{k})$ is given as:

$\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^{2}+6^{2}+(-2)^{2}}}=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{4+4 \lambda+\lambda^{2}+36+4}}=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^{2}+4 \lambda+44}}$

Scalar product of $(\hat{i}+\hat{j}+\hat{k})$ with this unit vector is 1 .'

$\Rightarrow(\hat{i}+\hat{j}+\hat{k}) \cdot \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^{2}+4 \lambda+44}}=1$

$\Rightarrow \frac{(2+\lambda)+6-2}{\sqrt{\lambda^{2}+4 \lambda+44}}=1$

$\Rightarrow \sqrt{\lambda^{2}+4 \lambda+44}=\lambda+6$

$\Rightarrow \lambda^{2}+4 \lambda+44=(\lambda+6)^{2}$

$\Rightarrow \lambda^{2}+4 \lambda+44=\lambda^{2}+12 \lambda+36$

$\Rightarrow 8 \lambda=8$

$\Rightarrow \lambda=1$

Hence, the value of λ is 1.