$\begin {array} {1 1} (A)\;k=3 & \quad (B)\;k=-3 \\ (C)\;k=\large\frac{1}{3} & \quad (D)\;k=-\large\frac{1}{3} \end {array}$

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- If two lines are parallel then their slopes are equal.
- If the given line is parallel to $x$ - axis then its slope is zero.

The given equation of line is

$(k-3)x-(4-k^2)y+k^2-7k+6=0$

If the given line is parallel to the $x$ - axis , the slope of the given line = slope of the $x$ - axis.

The given line can be written in the form of $y=mx+c$

(i.e.,) $y=\large\frac{(k-3)}{(4-k^2)}$$x+\large\frac{k^2-7k+6}{(4-k^2)}$

Hence the slope of the given line $ = \large\frac{(k-3)}{(4-k^2)}$

Slope of the $x$ - axis = 0

$ \therefore \large\frac{k-3}{4-k^2}=0$

$ \Rightarrow k-3=0$

$ \therefore k=3$

Hence the value of $ k = 3$.

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