The problem involves finding the intersection of two surfaces and determining the point on this intersection that is at a minimum distance from the \(xy\)-plane. Let's solve it step by step.
1. **Find the equations for the surfaces:**
- The first surface is z^2 = x^2 + y^2 which represents a double cone centered along the z-axis.
- The second surface is 4x + z = 7, which is a plane.
2. **Substitute to find the curve of intersection:**
- From the plane equation, express z in terms of x: z = 7 - 4x.
- Substitute z in the cone's equation:
(7 - 4x)^2 = x^2 + y^2
3. **Simplify the resulting equation:**
- Expand the left side:(7 - 4x)^2 = 49 - 56x + 16x^2
- Equate and simplify:
49 - 56x + 16x^2 = x^2 + y^2
15x^2 - 56x + 49 = y^2
4. **Find the minimum distance from the xy-plane:**
- The minimum distance to the xy-plane is determined by minimizing |z|.
- Since z = 7 - 4x, minimize |7 - 4x|.
5. **Optimize to find x (and consequently z):**
- For z = 0, solve 7 - 4x = 0.
- Thus, x = \frac{7}{4}.
6. **Find y using the curve relation:**
- Substitute x = \frac{7}{4} in the simplified intersection equation: 15\left(\frac{7}{4}\right)^2 - 56\left(\frac{7}{4}\right) + 49 = y^2
- Solve for y^2 to get the specific value.
7. **Calculate the distance from the origin:**
- We have point (x, y, 0), find its distance from the origin: \sqrt{x^2 + y^2 + z^2}.
\text{Distance} = \frac{7\sqrt{2}}{5}
Thus, the distance of point \( P \) from the origin is \frac{7\sqrt{2}}{5}.