The direction ratios of normal to the plane through $(1,0,0),(0,1,0)$ which makes an angle $\frac{\pi }{4}$ with the plane $x+y=3$ are

Let$I, m, n$ be the $d.r\text{'}s$ of the normal.
The $d.r\text{'}e$ of the line through (1, 0, 0) and (0, 1, 0) are
-1, 1, 0.
This is perpendicular to the normal

$\therefore -l+m=0,l=m$

The angle between the lines with$d.r\text{'}s I, I, n$ and 1, 1, 0
that of $x+y=3$ is $\frac{\pi }{4}$

$\begin{array}{l}\therefore \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}=\frac{l+l}{\sqrt{\left(2l^2+n^2\right)2}}.\\ \therefore 2l^2+n^2=4l^2,n=\sqrt{2}l\end{array}$

The $\text{ d.r's }$are $l,m,n$ or $l,l,\sqrt{2}l$ i.e., $1,1,\sqrt{2}\text{. }$