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

Any plane through ( 1, 0, 0) is $\mathrm{a}(\mathrm{x}-1)+\mathrm{by}+\mathrm{cz}=0----\left(\mathrm{i}\right)$

It passes through (0, 1, 0). Therefore, $\mathrm{a}(0-1)+\mathrm{b}\left(1\right)+\mathrm{c}\left(0\right)=0\text{ or }-\mathrm{a}+\mathrm{b}=0----\left(\mathrm{ii}\right)$

(i) makes an angle of $\frac{\mathrm{\pi }}{4}$ with x + y=3,therefore

$\begin{array}{l}\cos \frac{\mathrm{\pi }}{4}=\frac{\mathrm{a}\left(1\right)+\mathrm{b}\left(1\right)+\mathrm{c}\left(0\right)}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}\sqrt{1+1+0}}\\ \mathrm{or} \frac{1}{\sqrt{2}}=\frac{\mathrm{a}+\mathrm{b}}{\sqrt{2}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\\ \mathrm{or} \mathrm{a}+\mathrm{b}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}\end{array}$

Squaring, we get 

$\begin{array}{l}{\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}={\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\\ \mathrm{or} 2\mathrm{ab}={\mathrm{c}}^2\text{ or }2{\mathrm{a}}^2={\mathrm{c}}^2\\ \mathrm{or} \mathrm{c}=\sqrt{2}\mathrm{a}\end{array}$

$\begin{array}{r}\mathrm{Hence} \mathrm{a}:\mathrm{b}:\mathrm{c}=\mathrm{a}:\mathrm{a}:\sqrt{2}\mathrm{a}\\ =1:1:\sqrt{2}\end{array}$