Question 142, In a right triangular prism $\mathrm{ABC}-\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}$, the base edge lengths and the side edge lengths are all 2, and $\mathrm{E}$ is the midpoint of $\mathrm{CC}_{1}$, then the distance from $\mathrm{C}_{1}$ to the plane $\mathrm{AB} \mathrm{B}_{1} \mathrm{E}$ is $\qquad$ -
Question 142, Solution Method One: As shown in the figure, according to the conditions, we have $V_{C_{1}-A E B_{1}}=\frac{V_{A B C-A_{1} B_{1} C_{1}}}{2}-V_{A-A_{1} B_{1} C_{1}}=$ $\frac{V_{A B C}-A_{1} B_{1} C_{1}}{6}=\frac{2 \times \frac{\sqrt{3} \times 2^{2}}{4}}{6}=\frac{\sqrt{3}}{3}$. Let the distance from point $C_{1}$ to the plane $A B_{1} E$ be $h$. Since $A B_{1}=2 \sqrt{2}, A E=B_{1} E=$ $\sqrt{5}$, we have $S_{\triangle A E B_{1}}=\sqrt{6}$. Therefore, $\frac{\sqrt{6} h}{3}=\frac{\sqrt{3}}{3}$, solving for $h$ gives $h=\frac{\sqrt{2}}{2}$.

Solution Method Two: Establish a spatial rectangular coordinate system, set $C_{1}(0,0,0), B_{1}(0,2,0), A(\sqrt{3}, 1,2)$, $E(0,0,1)$. Then the equation of the plane $A E B$ is $\left|\begin{array}{ccc}x & y-2 & z \\ \sqrt{3} & -1 & 2 \\ 0 & -2 & 1\end{array}\right|=0$, which is $3 x-\sqrt{3}(y-2)-2 \sqrt{3} z=0$. Thus, the distance from point $C_{1}$ to the plane $\mathrm{AB}_{1} \mathrm{E}$ is $\mathrm{d}=\frac{|3 \times 0+\sqrt{3}(0-2)-2 \sqrt{3} \times 0|}{\sqrt{3^{2}+(\sqrt{3})^{2}+(2 \sqrt{3})^{2}}}=\frac{\sqrt{2}}{2}$.