1. **Problem:** Form a differential equation of all planes which are at a constant distance $a$ from the origin. 2. **Step 1:** Recall the equation of a plane: $$Ax + By + Cz + D = 0$$ 3. **Step 2:** The distance $d$ of the plane from the origin is given by $$d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$$ 4. **Step 3:** Since the distance is constant and equal to $a$, we have $$\frac{|D|}{\sqrt{A^2 + B^2 + C^2}} = a$$ 5. **Step 4:** Without loss of generality, normalize the plane equation so that $$\sqrt{A^2 + B^2 + C^2} = 1$$, then $$|D| = a$$ 6. **Step 5:** The general form of the plane at distance $a$ from origin is $$Ax + By + Cz + D = 0$$ with $$A^2 + B^2 + C^2 = 1$$ and $$|D| = a$$. 7. **Step 6:** To form the differential equation, consider the plane parameters as functions of $x,y,z$ and eliminate constants. The family of planes can be written as $$z = px + qy + r$$ where $p,q,r$ are constants related to $A,B,C,D$. 8. **Step 7:** Using the distance condition and differentiating with respect to $x,y,z$ to eliminate constants leads to the differential equation of the family of planes at distance $a$ from origin. **Final answer:** The differential equation representing all planes at constant distance $a$ from origin is $$\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 + 1 = \frac{z^2}{a^2}$$. This equation ensures the plane's normal vector magnitude and distance from origin are constant.