(i) Find the angle between the line :
` (x + 1)/(2) = (y)/(3) = (z - 3)/(6)` and the plane 10x + 12y - 11z = 3
(ii) Find the angle between the line :
`(x + 1)/(2) = (y -1)/(2) = (z -2)/(4)` and the plane 2x + y - 3z + 4 =0 .
(iii) Find the angle between the plane 2x + 4y - z = 8 and line `(x - 1)/(2) = (2 - y)/(7) = (3z + 6)/(12)`
(iv) Find the angle between the line `(x - 1)/(3) = (3 -y)/(-1) = (3z + 1)/(6)` and the plane 3x - 5y + 2z = 10 .

Answer

To solve the given problems step by step, we will use the formula for finding the angle between a line and a plane. The formula for the sine of the angle \( \theta \) between a line and a plane is given by:

\[ \sin \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

Where:
- \( (a_1, b_1, c_1) \) are the direction ratios of the line.
- \( (a_2, b_2, c_2) \) are the coefficients of the plane equation.

Let's solve each part of the question:

### (i) Find the angle between the line \( \frac{x + 1}{2} = \frac{y}{3} = \frac{z - 3}{6} \) and the plane \( 10x + 12y - 11z = 3 \).

1. **Identify direction ratios of the line**:
- From the line equation, we have \( a_1 = 2, b_1 = 3, c_1 = 6 \).

2. **Identify coefficients of the plane**:
- The plane equation can be written as \( 10x + 12y - 11z - 3 = 0 \), so \( a_2 = 10, b_2 = 12, c_2 = -11 \).

3. **Calculate \( a_1 a_2 + b_1 b_2 + c_1 c_2 \)**:
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 2 \cdot 10 + 3 \cdot 12 + 6 \cdot (-11) = 20 + 36 - 66 = -10 \]

4. **Calculate \( \sqrt{a_1^2 + b_1^2 + c_1^2} \)**:
\[ \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

5. **Calculate \( \sqrt{a_2^2 + b_2^2 + c_2^2} \)**:
\[ \sqrt{10^2 + 12^2 + (-11)^2} = \sqrt{100 + 144 + 121} = \sqrt{365} \]

6. **Calculate \( \sin \theta \)**:
\[ \sin \theta = \frac{|-10|}{7 \cdot \sqrt{365}} = \frac{10}{7\sqrt{365}} \]

7. **Find \( \theta \)**:
\[ \theta = \sin^{-1}\left(\frac{10}{7\sqrt{365}}\right) \]