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(D) Let the equation of the plane be $ax + by + cz = d$. Since it passes through $(0, -1, 0)$ and $(0, 0, 1)$,we have:$-b = d$ and $c = d$. Thus,$d =

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Both $(B)$ and $(C)$

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(D) Let the equation of the plane be $ax + by + cz = d$. Since it passes through $(0, -1, 0)$ and $(0, 0, 1)$,we have:$-b = d$ and $c = d$. Thus,$d = -b = c$.The equation becomes $ax - dy - dz = d$,or $ax - dy - dz - d = 0$.The normal vector is $\vec{n} = (a, -d, -d)$.The direction ratios of the line joining $(0, -1, 0)$ and $(0, 0, 1)$ are $(0, 1, 1)$. Since the line lies in the plane,the normal is perpendicular to it: $a(0) + (-d)(1) + (-d)(1) = 0 \Rightarrow -2d = 0$,which implies $d=0$ is not possible for a unique plane. Let's use $a, b, c$ directly.Points $A(0, -1, 0)$ and $B(0, 0, 1)$. Vector $\vec{AB} = (0, 1, 1)$.Normal $\vec{n} = (a, b, c)$. Since $\vec{AB}$ is in the plane,$\vec{n} \cdot \vec{AB} = 0 \Rightarrow b + c = 0 \Rightarrow c = -b$.Normal $\vec{n} = (a, b, -b)$.The angle between the plane and $y - z + 5 = 0$ is $\frac{\pi}{4}$. The normal to the second plane is $\vec{n_2} = (0, 1, -1)$.$\cos(\frac{\pi}{4}) = \frac{|\vec{n} \cdot \vec{n_2}|}{|\vec{n}| |\vec{n_2}|} \Rightarrow \frac{1}{\sqrt{2}} = \frac{|b + b|}{\sqrt{a^2 + b^2 + b^2} \cdot \sqrt{0^2 + 1^2 + (-1)^2}}$$\frac{1}{\sqrt{2}} = \frac{|2b|}{\sqrt{a^2 + 2b^2} \cdot \sqrt{2}} \Rightarrow \sqrt{a^2 + 2b^2} = 2|b|$Squaring both sides: $a^2 + 2b^2 = 4b^2 \Rightarrow a^2 = 2b^2 \Rightarrow a = \pm \sqrt{2}b$.If $a = \sqrt{2}b$,then $\vec{n} = (\sqrt{2}b, b, -b)$,which has direction ratios $(\sqrt{2}, 1, -1)$.If $a = -\sqrt{2}b$,then $\vec{n} = (-\sqrt{2}b, b, -b)$,which has direction ratios $(-\sqrt{2}, 1, -1)$.Option $(C)$ is $(\sqrt{2}, 1, -1)$. Option $(B)$ is $(2, \sqrt{2}, -\sqrt{2}) = \sqrt{2}(\sqrt{2}, 1, -1)$,which represents the same direction. Thus,both $(B)$ and $(C)$ are correct.

Find the direction ratios of the normal to the plane passing through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{\pi}{4}$ with the plane $y - z + 5 = 0$.

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