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The given equations are $\frac{2x-1}{4} = \frac{1-y}{3} = \frac{-z}{6}$. To find the direction ratios, we must express these in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Divide the numerator and denominator of each term to isolate the variables:
$\frac{2(x - 1/2)}{4} = \frac{-(y - 1)}{3} = \frac{z}{-6}$
$\frac{x - 1/2}{2} = \frac{y - 1}{-3} = \frac{z}{-6}$
From the standardized form, the direction ratios $(a, b, c)$ are $2, -3, -6$.
The magnitude of the direction vector is given by $\sqrt{a^2 + b^2 + c^2}$:
$\sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$
The direction cosines $(l, m, n)$ are calculated by dividing the direction ratios by the magnitude:
$l = \frac{2}{7}, m = \frac{-3}{7}, n = \frac{-6}{7}$
Final Answer: (B)