The equation of the plane passing through (1, | vedclass

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

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$x+y \pm \sqrt{2} z-1=0$

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(A) Let the equation of the plane be $a(x-1) + by + cz = 0$,which simplifies to $ax + by + cz - a = 0$. Since it passes through $(0,1,0)$,we have $a(0) + b(1) + c(0) - a = 0$,which implies $b = a$. So the equation is $ax + ay + cz - a = 0$,or $x + y + \frac{c}{a}z - 1 = 0$. Let $k = \frac{c}{a}$,then 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 between the planes is $45^{\circ}$,so $\cos(45^{\circ}) = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. $\frac{1}{\sqrt{2}} = \frac{|1+1+0|}{\sqrt{1^2+1^2+k^2} \sqrt{1^2+1^2+0^2}} = \frac{2}{\sqrt{2+k^2} \sqrt{2}}$. Squaring both sides: $\frac{1}{2} = \frac{4}{2(2+k^2)}$,which gives $2+k^2 = 4$,so $k^2 = 2$,$k = \pm \sqrt{2}$. Substituting $k$ back,we get $x + y \pm \sqrt{2}z - 1 = 0$.

The equation of the plane passing through $(1,0,0)$ and $(0,1,0)$ and making an angle of $45^{\circ}$ with the plane $x+y-3=0$ is:

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