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A quadric surface is defined by the equation x^2 - 2 x + y^2 - 6 y - 6 z^2 = 7. When this surface is rewritten in the form \frac{(x-x_0)^2}{a} + \frac{(y - y_0)^2}{b} + \frac{(z-z_0)^2}{c} = 1, what is the value of a?
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\[\begin{align*} x^2-2x+y^2-6y-6z^2&=7\\ \left(x^2-2x+1\right)+\left(y^2-6y+9\right)-6(z-0)^2&=7+1+9\\ (x-1)^2+(y-3)^2-6(z-0)^2&=17\\ \frac{(x-1)^2}{17}+\frac{(y-3)^2}{17}-\frac{6(z-0)^2}{17}&=1 \end{align*}\]
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