To solve this problem, we need to analyze the energy transformations as the particle falls.$\\$Initially, the particle is at height $S$ with potential energy and zero kinetic energy.$\\$The potential energy at height $h$ is given by:$\\ PE = mgh \\$The kinetic energy at height $h$ is given by:$\\ KE = \frac{1}{2} mv^2 \\$According to the problem, at a certain height, the kinetic energy is three times the potential energy:$\\ KE = 3 \cdot PE \\$Substitute the expressions for kinetic and potential energy:$\\ \frac{1}{2} mv^2 = 3 \cdot mgh \\$Cancel $m$ from both sides:$\\ \frac{1}{2} v^2 = 3gh \\$Solve for $v: \\ v^2 = 6gh \\ v = \sqrt{6gh} \\$Now, consider the conservation of mechanical energy. The total mechanical energy at height $S$ is:$\\ E_{total} = mgS \\$At height $h,$ the total mechanical energy is the sum of kinetic and potential energy:$\\ E_{total} = KE + PE \\ mgS = \frac{1}{2} mv^2 + mgh \\$Substitute $v^2 = 6gh$ into the equation:$\\ mgS = \frac{1}{2} m(6gh) + mgh \\ mgS = 3mgh + mgh \\ mgS = 4mgh \\$Cancel $mg$ from both sides:$\\ S = 4h \\$Solve for $h: \\ h = \frac{S}{4} \\$Now, substitute $h = \frac{S}{4}$ into $v = \sqrt{6gh}: \\ v = \sqrt{6g \cdot \frac{S}{4}} \\ v = \sqrt{\frac{6gS}{4}} \\ v = \sqrt{\frac{3gS}{2}} \\$Therefore, the height from the surface of Earth and the speed of the particle at that instant are:$\\ h = \frac{S}{4}, \quad v = \sqrt{\frac{3gS}{2}} \\$This corresponds to Option 4.