Step 1: Understanding the Question:
We need to find a plane equation given two points it passes through and its angle with another plane.
Step 3: Detailed Explanation:
Let the plane be \( ax + by + cz + d = 0 \).
1. Passes through \( (1, 0, 0) \implies a + d = 0 \implies d = -a \).
2. Passes through \( (0, 1, 0) \implies b + d = 0 \implies b = -d = a \).
So the plane is \( ax + ay + cz - a = 0 \), or \( x + y + \frac{c}{a}z - 1 = 0 \).
Let \( k = \frac{c}{a} \). Normal vector \( \vec{n}_1 = (1, 1, k) \).
3. Given plane \( x + y - 3 = 0 \) has normal \( \vec{n}_2 = (1, 1, 0) \).
4. Angle \( \theta = 45^\circ \):
\[ \cos 45^\circ = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|} \]
\[ \frac{1}{\sqrt{2}} = \frac{|1(1) + 1(1) + k(0)|}{\sqrt{1^2 + 1^2 + k^2}\sqrt{1^2 + 1^2 + 0^2}} = \frac{2}{\sqrt{2+k^2}\sqrt{2}} \]
\[ \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2+k^2}} \implies \sqrt{2+k^2} = 2 \]
\[ 2 + k^2 = 4 \implies k^2 = 2 \implies k = \pm \sqrt{2} \]
Substituting \( k \) back: \( x + y \pm \sqrt{2}z - 1 = 0 \).
Step 4: Final Answer:
The equation of the plane is \( x + y \pm \sqrt{2}z - 1 = 0 \).