1. **State the problem:** We need to determine the X-trics (characteristics) of the PDE given by $$z - p^2 = z$$ and find the particular integral surface passing through the parabola $$qpz + X^2 = 0$$ where $$f = z - p^2 + q^2 = 0$$. 2. **Rewrite the PDE:** Given $$z - p^2 = z$$, simplifying gives: $$z - p^2 - z = 0 \implies -p^2 = 0 \implies p^2 = 0$$ where $$p = \frac{\partial z}{\partial x}$$. 3. **Interpretation:** Since $$p^2 = 0$$, it implies $$p = 0$$, so $$\frac{\partial z}{\partial x} = 0$$. 4. **Use the Cauchy method of characteristics:** The PDE can be written as $$f = z - p^2 + q^2 = 0$$ where $$p = \frac{\partial z}{\partial x}$$ and $$q = \frac{\partial z}{\partial y}$$. 5. **Characteristic equations:** The characteristic system for first order PDEs is: $$\frac{dx}{f_p} = \frac{dy}{f_q} = \frac{dz}{p f_p + q f_q}$$ where $$f_p = \frac{\partial f}{\partial p}$$ and $$f_q = \frac{\partial f}{\partial q}$$. 6. **Calculate partial derivatives:** $$f = z - p^2 + q^2$$ $$f_p = -2p$$ $$f_q = 2q$$ 7. **Substitute into characteristic equations:** $$\frac{dx}{-2p} = \frac{dy}{2q} = \frac{dz}{p(-2p) + q(2q)} = \frac{dz}{-2p^2 + 2q^2}$$ 8. **Since from step 3, $$p=0$$, substitute $$p=0$$:** $$\frac{dx}{0} = \frac{dy}{2q} = \frac{dz}{2q^2}$$ 9. **Interpret the first ratio:** $$\frac{dx}{0}$$ implies $$dx=0$$, so $$x = \text{constant}$$ along characteristics. 10. **From the other ratios:** $$\frac{dy}{2q} = \frac{dz}{2q^2} \implies \frac{dy}{q} = \frac{dz}{q^2}$$ 11. **Rewrite:** $$\frac{dy}{q} = \frac{dz}{q^2} \implies dy = \frac{q}{q^2} dz = \frac{1}{q} dz$$ 12. **But $$q = \frac{\partial z}{\partial y}$$, so this is consistent with the definition of $$q$$. 13. **Integrate characteristic equations:** Since $$x = c_1$$ (constant), characteristics lie on planes parallel to the yz-plane. 14. **Use the parabola condition:** The parabola is given by $$qpz + X^2 = 0$$. Since $$p=0$$, this reduces to: $$q \cdot 0 \cdot z + X^2 = 0 \implies X^2 = 0 \implies X=0$$ 15. **Therefore, the particular integral surface passing through the parabola is on the plane $$x=0$$.** 16. **Final answer:** The X-trics are planes $$x = \text{constant}$$, and the particular integral surface passing through the parabola is the plane $$x=0$$.