Find the angle between the lines `(x+1)/(2)=(y)/(3)=(z-3)/(6)` and the planes `3x+y+z=7`.

Answer

To find the angle between the line given by the equation \(\frac{x+1}{2} = \frac{y}{3} = \frac{z-3}{6}\) and the plane defined by the equation \(3x + y + z = 7\), we will follow these steps:

### Step 1: Identify the direction ratios of the line
The line can be expressed in the symmetric form:
\[ \frac{x + 1}{2} = \frac{y}{3} = \frac{z - 3}{6} \]
From this, we can extract the direction ratios of the line:
- \(a = 2\)
- \(b = 3\)
- \(c = 6\)

### Step 2: Identify the normal vector of the plane
The equation of the plane is given as:
\[ 3x + y + z = 7 \]
From this, we can identify the coefficients of \(x\), \(y\), and \(z\) as the components of the normal vector to the plane:
- \(A = 3\)
- \(B = 1\)
- \(C = 1\)

### Step 3: Use the formula for the sine of the angle between a line and a plane
The formula for the sine of the angle \(\phi\) between a line and a plane is given by:
\[ \sin \phi = \frac{|A \cdot a + B \cdot b + C \cdot c|}{\sqrt{A^2 + B^2 + C^2} \cdot \sqrt{a^2 + b^2 + c^2}} \]

### Step 4: Substitute the values into the formula
Now substituting the values we have:
\[ \sin \phi = \frac{|3 \cdot 2 + 1 \cdot 3 + 1 \cdot 6|}{\sqrt{3^2 + 1^2 + 1^2} \cdot \sqrt{2^2 + 3^2 + 6^2}} \]

Calculating the numerator:
\[ 3 \cdot 2 + 1 \cdot 3 + 1 \cdot 6 = 6 + 3 + 6 = 15 \]

Calculating the denominator:
- For the plane:
\[ \sqrt{3^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \]
- For the line:
\[ \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

Thus, the denominator becomes:
\[ \sqrt{11} \cdot 7 = 7\sqrt{11} \]

### Step 5: Final calculation of \(\sin \phi\)
Putting it all together:
\[ \sin \phi = \frac{15}{7\sqrt{11}} \]

### Step 6: Find the angle \(\phi\)
To find the angle \(\phi\), we take the inverse sine:
\[ \phi = \sin^{-1}\left(\frac{15}{7\sqrt{11}}\right) \]

This gives us the angle between the line and the plane.