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By Cauchy-Schwarz, $$(9+16+144)(x^2+y^2+z^2)\ge (3x+4y+12z{)}^2.$$Since $x^2+y^2+z^2=1,$ this gives us $(3x+4y+12z{)}^2\le 169.$ Therefore, $3x+4y+12z\le 13.$

For equality to occur, we must have $$\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{144}.$$Since we want the maximum value of $3x+4y+12z,$ we can assume that $x,$ $y,$ and $z$ are all positive. Hence, $$\frac{x}{3}=\frac{y}{4}=\frac{z}{12}.$$Let $k=\frac{x}{3}=\frac{y}{4}=\frac{z}{12}.$ Then $x=3k,$ $y=4k,$ and $z=12k.$ Substituting into $x^2+y^2+z^2=1,$ we get $$9k^2+16k^2+144k^2=1,$$so $169k^2=1.$ Since $k$ is positive, $k=\frac{1}{13}.$ Then $x=\frac{3}{13},$ $y=\frac{4}{13},$ and $z=\frac{12}{13}.$

Thus, equality is possible, so the maximum value is $\boxed{13}.$