Let α → = 4 i ^ + 3 j ^ + 5 k ^ and β → = i ^ + 2 j ^ - 4 k ^ . Let β → 1…

Given that $\overrightarrow{\alpha }=4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+5\hat{\mathrm{k}}$ and $\overrightarrow{\beta }=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-4\hat{\mathrm{k}}$ and $\overrightarrow{\beta }=\overrightarrow{{\beta }_1}+\overrightarrow{{\beta }_2}$

$\Rightarrow \overrightarrow{{\beta }_2}=\overrightarrow{\beta }-\overrightarrow{{\beta }_1} ..........\left(1\right)$

since ${\overrightarrow{\beta }}_1$ is parallel to $\overrightarrow{\alpha }$, $\overrightarrow{{\beta }_1}=t\overrightarrow{\alpha }$

$\Rightarrow \overrightarrow{{\beta }_1}=t\left(4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+5\hat{\mathrm{k}}\right)=4t\hat{\mathrm{i}}+3t\hat{\mathrm{j}}+5t\hat{\mathrm{k}} .........\left(2\right)$

Substituting the values of $\overrightarrow{{\beta }_1}$ and $\overrightarrow{\alpha }$ in $\left(1\right)$, we get

$\overrightarrow{{\beta }_2}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-4\hat{\mathrm{k}}-\left(4t\hat{\mathrm{i}}+3t\hat{\mathrm{j}}+5t\hat{\mathrm{k}}\right)=\left(1-4t\right)\hat{\mathrm{i}}+\left(2-3t\right)\hat{\mathrm{j}}+\left(-4-5t\right)\hat{\mathrm{k}} ............\left(3\right)$

since $\overrightarrow{{\beta }_2}$ is perpendicular to $\overrightarrow{\alpha }$, so $\overrightarrow{{\beta }_2}·\overrightarrow{\alpha }=0$

$\Rightarrow \left[\left(1-4t\right)\hat{\mathrm{i}}+\left(2-3t\right)\hat{\mathrm{j}}+\left(-4-5t\right)\hat{\mathrm{k}} \right]·\left[4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+5\hat{\mathrm{k}}\right]=0$

$\Rightarrow 4\left(1-4t\right)+3\left(2-3t\right)+5\left(-4-5t\right)=0$

$\Rightarrow 4-16t+6-9t-20-25t=0$

$\Rightarrow -50t=10$$\Rightarrow t=\frac{-1}{5}$

From $\left(2\right)$ and $\left(3\right)$, we get  $\overrightarrow{{\beta }_1}=\frac{-1}{5}\left(4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+5\hat{\mathrm{k}}\right)$

$\overrightarrow{{\beta }_2}=\frac{9}{5}\hat{\mathrm{i}}+\frac{13}{5}\hat{\mathrm{j}}-3\hat{\mathrm{k}}=\frac{1}{5}\left(9\hat{\mathrm{i}}+13\hat{\mathrm{j}}-15\hat{\mathrm{k}}\right)$

So, the value of $5{\overrightarrow{\beta }}_2·\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)$ will be,

$=5\times \frac{1}{5}(9\hat{\mathrm{i}}+13\hat{\mathrm{j}}-15\hat{\mathrm{k}})·\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)=9+13-15=7$