The equation of plane passing through $(1,0,0)$ and $(0,1,0)$ and making an angle $45^{\circ}$ with the…

The required plane passes through $(1, 0, 0)$ and $(0, 1, 0)$, and makes a $45^\circ$ angle with the plane $x + y - 3 = 0$. Let the general equation be $Ax + By + Cz + D = 0$.

Substituting $(1, 0, 0)$ gives $A + D = 0 \Rightarrow A = -D$.
Substituting $(0, 1, 0)$ gives $B + D = 0 \Rightarrow B = -D$.

Using these, the equation becomes $-Dx - Dy + Cz + D = 0$.
For $D \neq 0$, divide by $D$: $-x - y + \frac{C}{D}z + 1 = 0$.
Let $k = \frac{C}{D}$, so the plane becomes $x + y - kz - 1 = 0$.

The normal vector is $\vec{n}_1 = (1, 1, -k)$.
The given plane $x + y - 3 = 0$ has normal vector $\vec{n}_2 = (1, 1, 0)$.

The angle $\theta$ between planes satisfies $\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{\|\vec{n}_1\| \|\vec{n}_2\|}$.
Given $\theta = 45^\circ$, so $\cos45^\circ = \frac{1}{\sqrt{2}}$.

Compute:
$\vec{n}_1 \cdot \vec{n}_2 = 1 \cdot 1 + 1 \cdot 1 + (-k) \cdot 0 = 2$
$\|\vec{n}_1\| = \sqrt{1^2 + 1^2 + (-k)^2} = \sqrt{2 + k^2}$
$\|\vec{n}_2\| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$

Substitute into the formula:
$\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2 + k^2} \cdot \sqrt{2}} = \frac{2}{\sqrt{2(2 + k^2)}}$

Square both sides:
$\left(\frac{1}{\sqrt{2}}\right)^2 = \left(\frac{2}{\sqrt{2(2 + k^2)}}\right)^2 \Rightarrow \frac{1}{2} = \frac{4}{2(2 + k^2)} \Rightarrow \frac{1}{2} = \frac{2}{2 + k^2}$

Cross-multiplying: $2 + k^2 = 4 \Rightarrow k^2 = 2 \Rightarrow k = \pm\sqrt{2}$

Substitute back into the plane equation:
$x + y - (\sqrt{2})z - 1 = 0$ or $x + y + \sqrt{2}z - 1 = 0$
Thus, the required plane is $x + y \pm \sqrt{2}z - 1 = 0$, corresponding to option A.