Thank you for reporting, we will resolve it shortly
Q.
The direction ratios of normal to the plane through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{\pi}{4}$ with the plane $y-z+5=0$ are :
Let the equation of plane be $a(x - 0) + b(y + 1) + c(z - 0) = 0$
It passes through $(0,0,1)$ then $b + c = 0$ .....(1)
Now $\cos \frac{\pi}{4} = \frac{a\left(0\right)+b\left(1\right)+c\left(-1\right)}{\sqrt{2} \sqrt{a^{2}+b^{2}+c^{2}}} $
$\Rightarrow a^{2} = - 2bc$ and $ b=-c $
we get $a^{2} = 2c^{2} $
$\Rightarrow a = \pm \sqrt{2}c $
$\Rightarrow $ direction ratio (a, b, c) = $(\sqrt{2} , - 1 , 1)$ or $(\sqrt{2} , 1 , -1)$