The sine of the angle between the straight line `(x - 2)/(3) = (y - 3)/(4) = (z - 4)/(5)` and the plane `2x - 2y + z = 5` is

Options

A. `(10)/(6sqrt5)`
B. `(4)/(5sqrt2)`
C. `(2sqrt3)/(5)`
D. `(sqrt2)/(10)`

Answer

To find the sine of the angle between the straight line given by the equations \((x - 2)/3 = (y - 3)/4 = (z - 4)/5\) and the plane given by the equation \(2x - 2y + z = 5\), we can follow these steps:

### Step-by-Step Solution

1. **Identify the direction ratios of the line**:
The line can be expressed in parametric form as:
\[ x = 2 + 3t, \quad y = 3 + 4t, \quad z = 4 + 5t \]
From this, the direction ratios of the line are \(a = 3\), \(b = 4\), \(c = 5\).

2. **Identify the normal vector of the plane**:
The equation of the plane is given as \(2x - 2y + z = 5\). The coefficients of \(x\), \(y\), and \(z\) give the normal vector of the plane:
\[ \vec{n} = (2, -2, 1) \]

3. **Calculate the angle between the line and the normal vector**:
The cosine of the angle \(\theta\) between the line and the normal vector can be calculated using the dot product formula:
\[ \cos \theta = \frac{\vec{a} \cdot \vec{n}}{|\vec{a}| |\vec{n}|} \]
where \(\vec{a} = (3, 4, 5)\) and \(\vec{n} = (2, -2, 1)\).

4. **Calculate the dot product \(\vec{a} \cdot \vec{n}\)**:
\[ \vec{a} \cdot \vec{n} = 3 \cdot 2 + 4 \cdot (-2) + 5 \cdot 1 = 6 - 8 + 5 = 3 \]

5. **Calculate the magnitudes \(|\vec{a}|\) and \(|\vec{n}|\)**:
\[ |\vec{a}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]
\[ |\vec{n}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

6. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{3}{(5\sqrt{2})(3)} = \frac{3}{15\sqrt{2}} = \frac{1}{5\sqrt{2}} \]

7. **Find \(\sin \theta\)**:
Since \(\theta\) is the angle between the line and the normal vector, the angle between the line and the plane is \(90^\circ - \theta\). Thus, we have:
\[ \sin(90^\circ - \theta) = \cos \theta \]
Therefore,
\[ \sin \theta = \frac{1}{5\sqrt{2}} \]

8. **Rationalize the sine value**:
To rationalize \(\sin \theta\):
\[ \sin \theta = \frac{1}{5\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{10} \]

### Final Answer
The sine of the angle between the straight line and the plane is:
\[ \sin \theta = \frac{\sqrt{2}}{10} \]