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(C) Let the equation of the plane be $a(x-1) + by + cz = 0$,which simplifies to $ax + by + cz = a$. Since it passes through $(0, 1, 0)$,we have $a(0)

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$(1, 1, \sqrt{2})$

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(C) Let the equation of the plane be $a(x-1) + by + cz = 0$,which simplifies to $ax + by + cz = a$. Since it passes through $(0, 1, 0)$,we have $a(0) + b(1) + c(0) = a$,so $b = a$. The equation becomes $ax + ay + cz = a$,or $x + y + \frac{c}{a}z = 1$. Let $k = \frac{c}{a}$. The normal vector is $\vec{n_1} = (1, 1, k)$. The normal to the plane $x + y - 3 = 0$ is $\vec{n_2} = (1, 1, 0)$. The angle $	heta = \frac{\pi}{4}$ between the planes is given by $\cos(\frac{\pi}{4}) = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. Thus,$\frac{1}{\sqrt{2}} = \frac{|1+1+0|}{\sqrt{1^2+1^2+k^2} \sqrt{1^2+1^2+0^2}} = \frac{2}{\sqrt{2+k^2} \sqrt{2}}$. Squaring both sides,$\frac{1}{2} = \frac{4}{2(2+k^2)}$,which implies $2+k^2 = 4$,so $k^2 = 2$ and $k = \pm \sqrt{2}$. For $k = \sqrt{2}$,the normal is $(1, 1, \sqrt{2})$.

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

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