Question:

Step 1: Let the direction ratios of the normal to the required plane be $(l, m, n)$. Step 2: The plane passes through the points $(0, -1, 0)$ and $(0, 0, 1)$. The direction ratios of the line segment connecting these points are $(0-0, 0-(-1), 1-0) = (0, 1, 1)$. Since the normal to the plane is perpendicular to any line lying in the plane, the dot product of the normal's direction ratios and the line's direction ratios must be zero: $$l(0) + m(1) + n(1) = 0$$ $$m + n = 0 \implies n = -m$$ Step 3: The given plane is $y - z + 5 = 0$. The direction ratios of its normal are $(0, 1, -1)$. Step 4: The angle between the two planes is $\frac{\pi}{4}$. The cosine of the angle between two planes with normals $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by: $$\cos(\theta) = \frac{|l_1 l_2 + m_1 m_2 + n_1 n_2|}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}}$$ Substituting the given values: $$\cos\left(\frac{\pi}{4}\right) = \frac{|l(0) + m(1) + n(-1)|}{\sqrt{l^2+m^2+n^2}\sqrt{0^2+1^2+(-1)^2}}$$ $$\frac{1}{\sqrt{2}} = \frac{|m - n|}{\sqrt{l^2+m^2+n^2}\sqrt{2}}$$ Step 5: Substitute $n = -m$ into the equation from Step 4: $$\frac{1}{\sqrt{2}} = \frac{|m - (-m)|}{\sqrt{l^2+m^2+(-m)^2}\sqrt{2}}$$ $$\frac{1}{\sqrt{2}} = \frac{|2m|}{\sqrt{l^2+2m^2}\sqrt{2}}$$ Multiplying both sides by $\sqrt{2}$: $$1 = \frac{|2m|}{\sqrt{l^2+2m^2}}$$ $$\sqrt{l^2+2m^2} = |2m|$$ Squaring both sides: $$l^2+2m^2 = (2m)^2$$ $$l^2+2m^2 = 4m^2$$ $$l^2 = 2m^2$$ $$l = \pm\sqrt{2}m$$ Step 6: From $n = -m$ and $l = \pm\sqrt{2}m$, the direction ratios $(l, m, n)$ can be expressed. Let $m = k$ for any non-zero scalar $k$. Then $n = -k$ and $l = \pm\sqrt{2}k$. Thus, the direction ratios of the normal to the required plane are $(\pm\sqrt{2}k, k, -k)$. For instance, choosing $k=1$, the direction ratios are $(\pm\sqrt{2}, 1, -1)$. Choosing $k=-1$, the direction ratios are $(\mp\sqrt{2}, -1, 1)$.