The sine of the angle between the straight line $\frac{x-3}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and the plane #2x- 2y + z = 5, $ is

$\frac{\sqrt{2}}{5}$

The line is parallel to the vector $\vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ and the plane is normal to the vector $\vec{n} = 2\hat{i} - 2\hat{j} + \hat{k}.$

If θ is the  angle between the line and the plane. Then, 

$sin θ=\frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|}⇒ sin \, \theta = \frac{6-8+5}{\sqrt{50}\sqrt{9}}=\frac{1}{5\sqrt{2}}=\frac{\sqrt{2}}{10}.$