Let's break down the problem step by step and address the different parts.

(a) Finding the lengths of $BD$, $ED$, and $AE$

• $BD$: Since $ABCD$ is a square with side length 3 m, the diagonal $BD$ can be found using the Pythagorean theorem in $\triangle ABD$: $BD = \sqrt{AB^2 + AD^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \, \text{m}.$

• $ED$: $ED$ is a vertical distance, where $E$ is directly below $B$ on the plane II, and $B$ is 2 meters above plane II. Since $BD = 3\sqrt{2}$ and $BE = 2$ (the height of the folded sheet), we can use the Pythagorean theorem in $\triangle BED$: $ED = \sqrt{BD^2 - BE^2} = \sqrt{(3\sqrt{2})^2 - 2^2} = \sqrt{18 - 4} = \sqrt{14} \, \text{m}.$

• $AE$: $AE$ is the distance from point $A$ to point $E$, and since $A$ is at ground level and $E$ is directly below $B$, we can treat this as a 2D problem. In triangle $AEB$, using Pythagoras: $AE = \sqrt{AB^2 - BE^2} = \sqrt{3^2 - 2^2} = \sqrt{9 - 4} = \sqrt{5} \, \text{m}.$

(b) Finding $\angle LADE$

We need to find the angle $\angle LADE$, where $L$ is the fold line and $A$, $D$, and $E$ form a triangle in plane II. This is a geometrical problem where $\triangle ADE$ has known side lengths, so we can use trigonometry.

First, use the dot product or cosine rule for the angle: $\cos \theta = \frac{AD^2 + AE^2 - ED^2}{2 \times AD \times AE}.$ Substitute the values: $\cos \theta = \frac{3^2 + \sqrt{5}^2 - \sqrt{14}^2}{2 \times 3 \times \sqrt{5}} = \frac{9 + 5 - 14}{6\sqrt{5}} = \frac{0}{6\sqrt{5}} = 0.$ Thus, $\theta = 90^\circ$, meaning $\angle LADE = 90^\circ$.

(c) The angle between $BD$ and the plane II

The angle between the line $BD$ and the plane II can be found using the sine of the angle. We know that the height $BE = 2$, and the length of $BD = 3\sqrt{2}$: [ \sin \theta = \frac{BE