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

For the line, the direction ratios are $3,4,5$, so a direction vector is $\vec{d}=(3,4,5)$. For the plane $2x-2y+z=5$, a normal vector is $\vec{n}=(2,-2,1)$.<br><br>If $\theta$ is the angle between the line and the plane, then $$\sin\theta=\frac{|\vec{d}\cdot \vec{n}|}{|\vec{d}|\,|\vec{n}|}.$$ Now, $$\vec{d}\cdot \vec{n}=3\cdot 2+4\cdot(-2)+5\cdot 1=3.$$ Also, $$|\vec{d}|=\sqrt{3^2+4^2+5^2}=\sqrt{50}=5\sqrt{2}.$$ And, $$|\vec{n}|=\sqrt{2^2+(-2)^2+1^2}=3.$$ Therefore, $$\sin\theta=\frac{3}{(5\sqrt{2})(3)}=\frac{1}{5\sqrt{2}}=\frac{\sqrt{2}}{10}.$$ Comparing with the given options, this matches option $\frac{\sqrt{2}}{10}$.