If the lines $$\frac{x}{1} = \frac{y}{2} = \frac{z}{3},\,\frac{{x - 1}}{3} = \frac{{y - 2}}{{ - 1}} = \frac{{z - 3}}{4}$$         and $$\frac{{x + k}}{3} = \frac{{y - 1}}{2} = \frac{{z - 2}}{h}$$      are concurrent then :

Solution :
Any point on the first line is $$\left( {r,\,2r,\,3r} \right),$$   and any point on the second line is $$\left( {1 + 3r',\,2 - r',\,3 + 4r'} \right).$$      For the point of intersection, $$r = 1 + 3r',\,2r = 2 - r',\,3r = 3 + 4r'\,\,\,\, \Rightarrow r = 1{\text{ and }}r' = 0.$$
So, the point of intersection of the first two lines is $$\left( {1,\,2,\,3} \right).$$   It is on the third line. So, $$\frac{{1 + k}}{3} = \frac{{2 - 1}}{2} = \frac{{3 - 2}}{h}.$$