\(To determine the surface area of solid $CXYZ,$ we determine the area of each of the four triangular faces and sum them. Areas of $\triangle CZX$ and $\triangle CZY:$ Each of these triangles is right-angled and has legs of lengths 6 and 8; therefore, the area of each is $\frac{1}{2}(6)(8)=24$. Area of $\triangle CXY:$ This triangle is equilateral with side length $6.$ We draw the altitude from $C$ to $M$ on $XY.$ Since $\triangle CXY$ is equilateral, then $M$ is the midpoint of $XY.$ Thus, $\triangle CMX$ and $\triangle CMY$ are $30^\circ$-$60^\circ$-$90^\circ$ triangles. Using the ratios from this special triangle,$$CM = \frac{\sqrt{3}}{2}(CX)=\frac{\sqrt{3}}{2}(6)=3\sqrt{3}.$$Since $XY = 6,$ the area of $\triangle CXY$ is$$\frac{1}{2}(6)(3\sqrt{3})=9\sqrt{3}.$$Area of $\triangle XYZ:$ We have $XY = 6$ and $XZ = YZ = 10$ and drop an altitude from $Z$ to $XY.$ Since $\triangle XYZ$ is isosceles, this altitude meets $XY$ at its midpoint, $M,$ and we have$$XM = MY = \frac{1}{2}(XY)=3.$$By the Pythagorean Theorem,\begin{align*} ZM &= \sqrt{ZX^2 - XM^2} \\ &= \sqrt{10^2-3^2} \\ &= \sqrt{91}. \end{align*}Since $XY = 6,$ the area of $\triangle XYZ$ is$$\frac{1}{2}(6)(\sqrt{91})=3\sqrt{91}.$$Finally, the total surface area of solid $CXYZ$ is$$24+24+9\sqrt{3}+3\sqrt{91}=\boxed{48+9\sqrt{3}+3\sqrt{91}}.$$\)
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It's probably 17
