1. **Stating the problem:** We have a cube [OABCDEFG] with edge length 3 in the coordinate system Oxyz. Points A, C, and E lie on the coordinate axes. The plane $\alpha$ intersects the cube forming the triangle [BDF]. The plane $\alpha$ is perpendicular to the spatial diagonal [OG] of the cube. We need to find the equation of the plane $\alpha$ in the form $ax + by + cz + d = 0$. 2. **Known points and vectors:** - Point $G = (3,3,3)$ is the opposite vertex of the cube from $O = (0,0,0)$. - The diagonal vector $\overrightarrow{OG} = (3,3,3)$. - Since $\alpha$ is perpendicular to $\overrightarrow{OG}$, the normal vector $\vec{n} = (a,b,c)$ of the plane $\alpha$ is parallel to $\overrightarrow{OG}$. 3. **Normal vector of the plane:** Because $\alpha$ is perpendicular to $\overrightarrow{OG}$, we have: $$\vec{n} = (a,b,c) = k(3,3,3)$$ for some scalar $k$. 4. **Using points on the plane:** The triangle [BDF] lies on the plane $\alpha$. Points B, D, and F are vertices of the cube. Coordinates: - $B = (3,0,0)$ - $D = (3,3,0)$ - $F = (0,3,3)$ 5. **Equation of the plane:** The plane equation is: $$3ax + 3by + 3cz + d = 0$$ Since $a = b = c$ (from step 3), let $a = b = c = m$, then: $$3m x + 3m y + 3m z + d = 0 \implies m(x + y + z) + \frac{d}{3} = 0$$ 6. **Using point B to find $d$:** Substitute $B = (3,0,0)$ into the plane equation: $$m(3 + 0 + 0) + \frac{d}{3} = 0 \implies 3m + \frac{d}{3} = 0$$ Multiply both sides by 3: $$9m + d = 0 \implies d = -9m$$ 7. **Final plane equation:** Substitute $d = -9m$ back: $$m(x + y + z) - 9m = 0$$ Divide both sides by $m$ (assuming $m \neq 0$): $$\cancel{m}(x + y + z) - 9\cancel{m} = 0$$ So the plane equation is: $$x + y + z - 9 = 0$$ **Answer:** The equation of the plane $\alpha$ is: $$x + y + z = 9$$