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By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$, $B(4,0,0)$, and $C(0,4,0)$, we find that the equation of plane $ABC$ is: $$\frac{x}{4}+\frac{y}{4}+\frac{z}{3}=1,$$ so $3x+3y+4z-12=0.$ The equation for the distance of a point $(a,b,c)$ to a plane $Ax+By+Cz+D=0$ is given by: $$\frac{|Aa+Bb+Cc+D|}{\sqrt{A^2+B^2+C^2}}.$$ Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where $a=b=c=0$) to the plane is given by: $$\frac{D}{\sqrt{A^2+B^2+C^2}}=\frac{12}{\sqrt{9+9+16}}=\frac{12}{\sqrt{34}}.$$ Since $\sqrt{34}<6$, this number should be just a little over $2$, and the correct answer is $\boxed{2.1}$. Note that the equation above for the distance from a point to a plane is a 3D analogue of the 2D case of the distance formula, where you take the distance from a point to a line. In the 2D case, both $c$ and $C$ are set equal to $0$.