Explanation:
C Given,
Point \((3,4,5)\)
line : \(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\)
Plane: \(\mathrm{x}+\mathrm{y}+\mathrm{z}=17\)
Let, \(\quad \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda\)
\(\Rightarrow \quad \mathrm{Q}(\lambda+3,2 \lambda+4,2 \lambda+5)\)
Now,
\(x+y+z=17\)
\(\lambda+3+2 \lambda+4+2 \lambda+5=17\)
\(\Rightarrow 5 \lambda=5\)
\(\Rightarrow \lambda=1\)
\(\therefore\) Point \(=\mathrm{Q}(4,6,7)\)
Now, \(\mathrm{P}(3,4,5)\), and \(\mathrm{Q}(4,6,7)\)
Distance between two point,
\(P Q =\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}\)
\(=\sqrt{(1)^2+(2)^2+(2)^2}\)
\(=\sqrt{1+4+4}=\sqrt{9}=3\)