The sine of the angle between the straight line \(\frac{x-2}{2}=\frac{y-3}{4}=\frac{4-z}{2}\) are the plane \(2 x-2 y+z=5\) is

We have, the equation of line as
\(\frac{x-2}{3}=\frac{y-3}{4}=\frac{4-z}{-5}\)
\(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)
This line is parallel to the vector \(\mathbf{b}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\)
Equation of plane is \(2 x-2 y+z=5\)
Normal to the plane is \(\mathbf{n}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Then, \(\sin \theta=\frac{\mathbf{b} \cdot \mathbf{n}}{|\mathbf{b}||\mathbf{n}|}\)
\(=\frac{|(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})|}{\sqrt{3^2+4^2+5^2} \sqrt{4+4+1}}\)
\(=\frac{|6-8+5|}{\sqrt{50} \sqrt{9}}=\frac{3}{15 \sqrt{2}}=\frac{1}{5 \sqrt{2}}\)