The points on the line $\frac{x+1}{1}=\frac{y+3}{3}=\frac{z-2}{-2},$  distance $\sqrt{\left(14\right)}$  from the point in which the line meets the plane $3x+4y+5z-5=0$  are:

$\frac{x+1}{1}=\frac{y+3}{3}=\lambda$

$\Rightarrow$  any point on this line is $P\left(\lambda -1,3\lambda -3,-2\lambda +2\right)$ .

If this point lies in the plane $3x+4y+5z-5=0,$

Then $3\lambda -3+12\lambda -10\lambda +10-5=0$

$\Rightarrow$  $5\lambda -10=0\Rightarrow \lambda =2$ .

$\therefore$  Point of intersection of the line and the plane is $A\left(1,3,-2\right)$ .

It is given that $AP=\sqrt{14}$ .

$\therefore {\left(\lambda -2\right)}^2+{\left(3\lambda -6\right)}^2+{\left(-2\lambda +4\right)}^2=14$

$\Rightarrow 14{\lambda }^2-56\lambda +56=14$

$\Rightarrow {\lambda }^2-4\lambda +3=0\Rightarrow \lambda =1,3$