The equation of the line which passes through the point $$\left( {1,\,1,\,1} \right)$$   and intersect the lines $$\frac{{x - 1}}{2} = \frac{{y - 2}}{3} = \frac{{z - 3}}{4}$$     and $$\frac{{x + 2}}{1} = \frac{{y - 3}}{2} = \frac{{z + 1}}{4}$$     is :

Solution :
Any line passing through the point $$\left( {1,\,1,\,1} \right)$$   is
$$\frac{{x - 1}}{a} = \frac{{y - 1}}{b} = \frac{{z - 1}}{c}......\left( {\text{i}} \right)$$
This line intersects the line
$$\frac{{x - 1}}{2} = \frac{{y - 2}}{3} = \frac{{z - 3}}{4}$$
$${\text{If }}a:b:c \ne 2:3:4$$
\[{\rm{and }}\left| \begin{array}{l} 1 - 1\,\,\,\,\,2 - 1\,\,\,\,\, 3 - 1\\ \,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\ \,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \end{array} \right| = 0\]
$$ \Rightarrow a - 2b + c = 0.....\left( {{\text{ii}}} \right)$$
Again, line $$\left( {\text{i}} \right)$$ intersects line
$$\frac{{x - \left( { - 2} \right)}}{1} = \frac{{y - 3}}{2} = \frac{{z - \left( { - 1} \right)}}{4}$$
$${\text{If }}a:b:c \ne 2:3:4$$
\[{\rm{and }}\left| \begin{array}{l} - 2 - 1\,\,\,\,\,3 - 1\,\,\,\,\, - 1 - 1\\ \,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\ \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \end{array} \right| = 0\]
$$ \Rightarrow 6a + 5b - 4c = 0.....\left( {{\text{iii}}} \right)$$
From $$\left( {{\text{ii}}} \right)$$ and $$\left( {{\text{iii}}} \right)$$ by cross multiplication, we have
$$\frac{a}{{8 - 5}} = \frac{b}{{6 + 4}} = \frac{c}{{5 + 12}}{\text{ or }}\frac{a}{3} = \frac{b}{{10}} = \frac{c}{{17}}$$
So, the required line is
$$\frac{{x - 1}}{3} = \frac{{y - 1}}{{10}} = \frac{{z - 1}}{{17}}.$$