Answer:
Given equation of plane is\[x+y+z=2\] ?(i) Equation of the line is \[\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \,\,(say)\] ?(ii) Then, any point P on the line is \[P(\lambda +3,\,\,2\lambda +4,\,\,2\lambda +5)\] ?(iii) 
Since, the point P lies on the given plane. So, it will satisfies the plane. \[\therefore \] \[(\lambda +3)+(2\lambda +4)+(2\lambda +5)=2\] \[\Rightarrow \] \[5\lambda +12=2\] \[\Rightarrow \] \[5\lambda =-\,10\] \[\Rightarrow \] \[\lambda =-\,2\] On putting the value of \[\lambda \] in Eq. (iii), we get Coordinates of \[P=(-\,2+3,\,\,-\,4+4,\,\,-\,4+5)=(1,\,\,0,\,\,1)\] Now, the distance between P(1, 0, 1) and (3, 4, 5) is \[|PQ|\,=\sqrt{{{(3-1)}^{2}}+{{(4-0)}^{2}}+{{(5-1)}^{2}}}\] \[[\because d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}}}]\] \[=\sqrt{{{(2)}^{2}}+{{(4)}^{2}}+{{(4)}^{2}}}=\sqrt{4+16+16}\] \[=\sqrt{36}=6\,\,\text{units}\]
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