Let $\vec{a}$ and $\vec{b}$ be the adjacent sides of a parallelogram, where
$\begin{aligned}
& \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}
\end{aligned}$
Let diagonal be $\overrightarrow{\mathrm{c}}$
$\begin{aligned}
\vec{c} & =\vec{a}+\vec{b} \\
\vec{c} & =2 \hat{i}-4 \hat{j}+5 \hat{k}+\hat{i}-2 \hat{j}-3 \hat{k} \\
& =3 \hat{i}-6 \hat{j}+2 \hat{k}
\end{aligned}$
$\text { Magnitude of } \begin{aligned}
\vec{c} & =\sqrt{3^2+(-6)^2+(2)^2} \\
& =\sqrt{49}=7
\end{aligned}$
$\therefore \quad$ Unit vector in direction of diagonal $\overrightarrow{\mathrm{c}}$ is
$\begin{aligned}
& =\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|} \\
& =\frac{1}{7}(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\
& =\frac{3}{7} \hat{\mathrm{i}}-\frac{6}{7} \hat{\mathrm{j}}+\frac{2}{7} \hat{\mathrm{k}}
\end{aligned}$