The point $$A\left( {3,\, - 2,\,4} \right)$$   is shifted parallel to the line $$\frac{x}{{\sqrt 3 }} = \frac{{y - 1}}{2} = \frac{{z + 1}}{3}$$     by a distance $$1$$. The coordinates of $$P$$ in the new position are :

Solution :
Direction cosines of the line are $$\frac{{\sqrt 3 }}{4},\,\frac{2}{4},\,\frac{3}{4}.$$   So, $$P = \left( {3 \pm 1.\frac{{\sqrt 3 }}{4}, - 2 \pm 1.\,\frac{2}{4},4 \pm 1.\,\frac{3}{4}} \right)$$
$$\therefore \,\,P = \left( {3 \pm \frac{{\sqrt 3 }}{4}, - 2 \pm \,\frac{1}{2},4 \pm \,\frac{3}{4}} \right)$$