The sides of a parallelogram are $2\hat i +4\hat j-5\hat k$ and $\hat i+2\hat j+3\hat k$, then the unit vector parallel to one of the diagonals is

$\frac{1}{7}(3\hat i+6\hat j-2\hat k)$

Let

$\vec a=2\hat i+4\hat j-5\hat k, \vec b =\hat i +2\hat j+3\hat k$.

The diagonals of the parallelogram are

$\vec p=\vec a+\vec b, \vec q=\vec b - \vec a$

$⇒\vec p=3\hat i+6\hat j-2\hat k,\vec q=-\hat i-2\hat j+8\hat k$

So, unit vectors along the diagonals are

$\frac{1}{7}(3\hat i+6\hat j-2\hat k)$ and $\frac{1}{\sqrt{69}}(-\hat i-2\hat j+8\hat k)$

Hence, option (1) is correct.