1. **Problem Statement:** We need to find the equation of a plane passing through point $P(2, -1, 4)$ and perpendicular to vector $\vec{n} = \langle 3, 2, -1 \rangle$. 2. **Formula for Plane Equation:** The general equation of a plane with normal vector $\vec{n} = \langle a, b, c \rangle$ passing through point $P(x_0, y_0, z_0)$ is: $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$ 3. **Substitute values:** $$3(x - 2) + 2(y + 1) - 1(z - 4) = 0$$ 4. **Simplify:** $$3x - 6 + 2y + 2 - z + 4 = 0$$ $$3x + 2y - z = 0$$ 5. **Point Classification:** Substitute $Q(5, 5, 5)$ into the plane equation: $$3(5) + 2(5) - 5 = 15 + 10 - 5 = 20$$ Since $20 > 0$, the packet is **Accepted** (above the plane). 6. **Distance from Point to Plane:** Distance formula from point $Q(x_1, y_1, z_1)$ to plane $ax + by + cz + d = 0$ is: $$\text{Distance} = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$ Rewrite plane equation as: $$3x + 2y - z + 0 = 0$$ Calculate distance: $$\frac{|3(5) + 2(5) - 1(5) + 0|}{\sqrt{3^2 + 2^2 + (-1)^2}} = \frac{|20|}{\sqrt{9 + 4 + 1}} = \frac{20}{\sqrt{14}} = \frac{20}{3.7417} \approx 5.35$$ 7. **Computer Science Insight:** The distance of a point from the decision boundary (plane) indicates the confidence of classification. A larger distance means the point is classified with higher confidence, while a distance close to zero means the point lies near the boundary and classification is less certain. **Final answers:** - Plane equation: $3x + 2y - z = 0$ - Point $Q$ is above the plane and packet is Accepted. - Distance of $Q$ from plane is approximately $5.35$ units.