Answer:
Option C
Explanation:
Condition for two lines are coplannar
$\begin{bmatrix}x_{1}-x_{2} & y_{1}-y_{2} &z_{1}-z_{2} \\l_{1}& m_{1}&n_{1} \\ l_{2}&m_{2}&n_{2} \end{bmatrix}=0$
where , (x1,y1,z1) and (x2,y2,z2) are the points lie ona line (i) and (ii) respectively and <l1,m1,n1 > and
< l2,,m2,n2 > are the direction cosine of the line (i) and (ii) respectively.
$\therefore$ $\begin{vmatrix}2-1 & 3-4 &4-5 \\1& 1&-k \\ k&2&1 \end{vmatrix}=0$
$\begin{vmatrix}1 & -1 &-1 \\1& 1&-k \\ k&2&1 \end{vmatrix}=0$
$\Rightarrow$ $1(1+2k)+(1+k^{2})-(2-k)=0$
$\Rightarrow$ $ k^{2}+2k+k=0$
$\Rightarrow $ $k^{2}+3k=0\Rightarrow k=0,-3$
If 0 appears in the denominator , then the correct way of representing the equation of straight line is
$\frac{x-2}{1}=\frac{y-3}{1};z=4$
