The given equation of line is
\((k-3) x - (4 - k^ 2 ) y + k ^2- 7k + 6 = 0 … (1)\)
(a) If the given line is parallel to the x-axis, then Slope of the given line = Slope of the x-axis
The given line can be written as
\((4- k^ 2 ) y = (k- 3) x + k^ 2 - 7k + 6 = 0\)
\(y=\frac{(k-3)}{(4-k^2)}x+\frac{k^2-7k+6}{(4-k^2)}\)
∴ Slope of the given line =\(\frac{(k-3)}{(4-k^2)},\) which is of the form \(y = mx + c. \)
Slope of the x-axis = 0
\(∴ \frac{(k-3)}{(4-k^2)}=0\)
\(⇒ k-3=0\)
\(⇒ k=3\)
Thus, if the given line is parallel to the x-axis, then the value of k is 3.
(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.
The slope of the given line is \(\frac{(k-3)}{(4-k^2)}\)
Now, \(\frac{(k-3)}{(4-k^2)}\) is undefined at \(k^2 = 4 \)
\(k^2 = 4 \)
\(⇒ k = ±2 \)
Thus, if the given line is parallel to the y-axis, then the value of k is \(±2\).
(c) If the given line is passing through the origin, then point (0, 0) satisfies the given equation of line.
\((k-3)(0)-(4-k^2)(0)+k^2-7k+6=0\)
\(k^2-7k+6=0\)
\((k-6)(k-1)=0\)
\(k=1\space or\space 6\)
Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.