The scalar product of the vector a → = i ^ + j ^ + k ^ with a unit vector along the sum of the vectors…

Given,

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k} ...\left(i\right)$

$\overrightarrow{b}=2\hat{i}+4\hat{j}-5\hat{k}$

$\overrightarrow{c}=\lambda \hat{i}+2\hat{j}+3\hat{k}$

Now,

$\overrightarrow{b}+\overrightarrow{c}=\left(\lambda +2\right)\hat{i}+6\hat{j}-2\hat{k}$

Let unit vector $\overrightarrow{b}$ be the vector along $\overrightarrow{b}+\overrightarrow{c}$

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

$\Rightarrow \overrightarrow{b}=\frac{(\lambda +2)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(\lambda +2{)}^2+40}}...\left(ii\right)$

Now,

 $\overrightarrow{a}·\overrightarrow{b}=1$

$\Rightarrow \left(\hat{i}+\hat{j}+\hat{k}\right). \left\{\frac{(\lambda +2)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(\lambda +2{)}^2+40}}\right\}=1$

$\Rightarrow \frac{(\lambda +2)+6-2}{\sqrt{(\lambda +2{)}^2+40}}=1$

$\Rightarrow \frac{(\lambda +2)+6-2}{\sqrt{(\lambda +2{)}^2+40}}=1$

$\Rightarrow \frac{\lambda +6}{\sqrt{(\lambda +2{)}^2+40}}=1$

$\Rightarrow \lambda +6=\sqrt{{\left(\lambda +2\right)}^2+40}$

Squaring both sides, we get

$\Rightarrow (\lambda +6{)}^2=(\lambda +2{)}^2+40$

$\Rightarrow {\lambda }^2+12\lambda +36= {\lambda }^2+4\lambda +4+40$

$\Rightarrow 12\lambda +36=4\lambda +44$

$\Rightarrow 8\lambda =8$

$=\lambda =1$.