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 proportional to

$1, 1, \sqrt{2}$

Let the direction ratios of the normal to the plane be proportional to a, b, c. Then, the equation of the plane is

$a(x-1) + b(y-0) + c(z-0) = 0 $   ................(i)

It passes through (0, 1, 0).

$a(-1) + b(1) + c(0) = 0 ⇒ a = b $ ..........(ii)

It is given that the plane (i) makes an angle $\frac{\pi}{4}$ with the plane x + y = 3.

$∴ cos\frac{\pi}{4}=\frac{a×1+b×1+c×0}{\sqrt{a^2+b^2+c^2}\sqrt{1+1}}$

$⇒ \frac{1}{\sqrt{2}}=\frac{a+b}{\sqrt{a^2+b^2+c^2}\sqrt{2}}$

$⇒\sqrt{a^2+b^2+c^2} = a+ b $

$⇒ a^2 + b^2 + c^2 = a^2 + b^2 + 2ab ⇒ c^2 = 2ab$ ...............(iii)

From (ii) and (iii), we have

$a : b : c = a : a : \sqrt{2}a = 1 : 1 : \sqrt{2}$