If `theta` the angle between the line `(x+1)/(3) = (y-1)/(2) = (z-2)/(4)` and the plane `2x + y-3z+ 4 =0,` then `64 cosec ^(2) theta ` is equal to :

Options

A. 1246
B. 5625
C. 2500
D. 1624

Answer

To solve the problem, we need to find the value of \( 64 \csc^2 \theta \), where \( \theta \) is the angle between the line given by the equations \(\frac{x+1}{3} = \frac{y-1}{2} = \frac{z-2}{4}\) and the plane defined by the equation \(2x + y - 3z + 4 = 0\).

### Step 1: Identify the Direction Vector of the Line
The line can be expressed in parametric form. From the equation \(\frac{x+1}{3} = \frac{y-1}{2} = \frac{z-2}{4}\), we can identify the direction vector of the line:

\[ \text{Direction vector } \mathbf{d} = \langle 3, 2, 4 \rangle \]

### Step 2: Identify the Normal Vector of the Plane
The plane equation \(2x + y - 3z + 4 = 0\) gives us the normal vector of the plane directly from the coefficients of \(x\), \(y\), and \(z\):

\[ \text{Normal vector } \mathbf{n} = \langle 2, 1, -3 \rangle \]

### Step 3: Calculate the Dot Product of the Normal and Direction Vectors
Next, we calculate the dot product of the normal vector \(\mathbf{n}\) and the direction vector \(\mathbf{d}\):

\[ \mathbf{n} \cdot \mathbf{d} = 2 \cdot 3 + 1 \cdot 2 + (-3) \cdot 4 = 6 + 2 - 12 = -4 \]

### Step 4: Calculate the Magnitudes of the Vectors
Now, we need to find the magnitudes of both vectors.

1. Magnitude of the normal vector \(\mathbf{n}\):

\[ |\mathbf{n}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]

2. Magnitude of the direction vector \(\mathbf{d}\):

\[ |\mathbf{d}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29} \]

### Step 5: Calculate \(\sin \theta\)
The sine of the angle \(\theta\) between the line and the plane can be calculated using the formula:

\[ \sin \theta = \frac{|\mathbf{n} \cdot \mathbf{d}|}{|\mathbf{n}| |\mathbf{d}|} \]

Substituting the values we found:

\[ \sin \theta = \frac{|-4|}{\sqrt{14} \cdot \sqrt{29}} = \frac{4}{\sqrt{14 \cdot 29}} \]

### Step 6: Calculate \(\csc^2 \theta\)
Since \(\csc \theta = \frac{1}{\sin \theta}\), we have:

\[ \csc^2 \theta = \frac{1}{\sin^2 \theta} = \frac{|\sqrt{14 \cdot 29}|}{16} \]

### Step 7: Calculate \(64 \csc^2 \theta\)
Now we can find \(64 \csc^2 \theta\):

\[ 64 \csc^2 \theta = 64 \cdot \frac{14 \cdot 29}{16} = 4 \cdot 14 \cdot 29 \]

Calculating \(4 \cdot 14 \cdot 29\):

\[ 4 \cdot 14 = 56 \]
\[ 56 \cdot 29 = 1624 \]

Thus, the final answer is:

\[ \boxed{1624} \]