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

It is given that, the equation of line as
$\frac{{x - 2}}{3} = \frac{{y - 3}}{4} = \frac{{z - 4}}{5}$

Now, the line passes through point (2,3,4) and having direction ratios (3,4,5) .

Since, the line passes through point (2,3,4) and parallel to the vector $(3\widehat {\rm{i}} + 4\widehat {\rm{j}} + 5\widehat {\rm{k}})$.

$\therefore$ $\overrightarrow {\rm{b}} = 3\widehat {\rm{i}} + 4\widehat {\rm{j}} + 5\widehat {\rm{k}}$

Also, the cartesian form of the given plane is $2x - 2y + z = 5$.

$\Rightarrow$ $(x\widehat {\rm{i}} + y\widehat {\rm{j}} + z\widehat {\rm{k}})(2\widehat {\rm{i}} - 2\widehat {\rm{j}} + \widehat {\rm{k}}) = 5$

$\therefore$ $\overrightarrow {\rm{n}} = (2\widehat {\rm{i}} - 2\widehat {\rm{j}} + \widehat {\rm{k}})$

As we know, $\sin \theta = \frac{{|\overrightarrow {\rm{b}} \cdot \overrightarrow {\rm{n}} |}}{{|\overrightarrow {\rm{b}} | \cdot |\overrightarrow {\rm{n}} |}}$

$= \frac{{|(3\widehat {\rm{i}} + 4\widehat {\rm{j}} + 5\widehat {\rm{k}}) \cdot (2\widehat {\rm{i}} - 2\widehat {\rm{j}} + \widehat {\rm{k}})|}}{{\sqrt {{3^2} + {4^2} + {5^2}} \cdot \sqrt {4 + 4 + 1} }}$

$= \frac{{|6 - 8 + 5|}}{{\sqrt {50} \cdot 3}} = \frac{3}{{15\sqrt 2 }} = \frac{1}{{5\sqrt 2 }}$
$\sin \theta = \frac{{\sqrt 2 }}{{10}}$