The scalar product of the vector $\widehat i + \widehat j + \widehat k$ with a unit vector along the sum of vectors$2\hat i + 4\hat j - 5\hat k$ and $\lambda \hat i + 2\widehat j + 3\hat k$ is equal to one. Find the value of $\lambda$.

Let $\overrightarrow a = \widehat i + \widehat j + \widehat k$, $\overrightarrow b = 2\hat i + 4\hat j - 5\hat k$ and $\overrightarrow c = \lambda \hat i + 2\widehat j + 3\hat k$

Then, $\vec b + \vec c = (2\hat i + 4\widehat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k) = (\lambda + 2)\hat i + 6\widehat j - 2\hat k$

Now, unit vector along$\vec b + \vec c = \cfrac{{(\lambda + 2)\hat i + 6\widehat j - 2\hat k}}{{\sqrt {{{(\lambda + 2)}^2} + 36 + 4} }}$.

By the question,
$(\widehat i + \widehat j + \widehat k) \cdot \cfrac{{((\lambda + 2)\hat i + 6\hat j - 2\hat k)}}{{\sqrt {{{(\lambda + 2)}^2} + 40} }} = 1 \Rightarrow \cfrac{{(\lambda + 2 + 6 - 2)}}{{\sqrt {{{(\lambda + 2)}^2} + 40} }} = 1$

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

Squaring both sides,

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

$\Rightarrow 8\lambda = 8 \Rightarrow \lambda = 1$