Let  $\overline{\alpha }=4\overline{i}+3\overline{j}+5\overline{k }\text{\hspace{0.17em}and }\overline{ \beta }=\overline{i}+2\overline{j}-4\overline{k}.$Let $\overline{{\beta }_1}$ is a parallel to $\overline{\alpha }\text{\hspace{0.17em} and }\overline{{\beta }_2}$ be perpendicular to $\overline{\alpha }$. If
$\overline{\beta }=\overline{{\beta }_1}+\overline{{\beta }_2}$  Then the value of $5\overline{{\beta }_2}.(\overline{i}+\overline{j}+\overline{k})\text{\hspace{0.17em}is}$

$$\begin{gathered} \overline{{\beta }_1}=t\overline{\alpha }\overline{\alpha }.\overline{{\beta }_2}=0 \\ \overline{\beta }=\overline{{\beta }_1}+\overline{{\beta }_2} \end{gathered}$$

Taking dot with  $\overline{\alpha }$

$\overline{\alpha }.\overline{\beta }=\overline{\alpha }.\overline{{\beta }_1}+\overline{\alpha }.\overline{{\beta }_2}$

$4+6-20=\overline{\alpha }.t\overline{\alpha }+0$

$-10=t{\left|\overline{\alpha }\right|}^2$

$-10=t(16+9+25)$

$\therefore t=\frac{-1}{5}$

$\therefore \overline{\beta }=-\frac{1}{5}\overline{\alpha }+{\beta }_2$

$$\begin{gathered} \therefore \overline{{\beta }_2}=\overline{\beta }+\frac{1}{5}\overline{\alpha } \\ =(\overline{i}+2\overline{j}-4\overline{k})+\frac{1}{5}(4\overline{i}+3\overline{j}+5\overline{k}) \\ =\frac{9\overline{i}+13\overline{j}-15\overline{k}}{5} \end{gathered}$$

$$\begin{gathered} \overline{5{\beta }_2}.(\overline{i}+\overline{j}+\overline{k})=(9\overline{i}+13\overline{j}-15\overline{k}).(\overline{i}+\overline{j}+\overline{k}) \\ =9+13-15=7 \end{gathered}$$