The direction cosines of the normal to the plane
`x + 2y - 3z + 4 = 0` are

Options

A. `(-1)/sqrt(14), (-2)/sqrt(14),3/sqrt(14)`
B. `(1)/sqrt(14), (2)/sqrt(14),3/sqrt(14)`
C. `(-1)/sqrt(14), (2)/sqrt(14),3/sqrt(14)`
D. `(1)/sqrt(14), (-2)/sqrt(14),(-3)/sqrt(14)`

Answer

To find the direction cosines of the normal to the plane given by the equation \( x + 2y - 3z + 4 = 0 \), we can follow these steps:

### Step 1: Identify the coefficients
The equation of the plane can be written in the standard form \( ax + by + cz + d = 0 \). From the given equation:
- \( a = 1 \)
- \( b = 2 \)
- \( c = -3 \)
- \( d = 4 \)

### Step 2: Calculate the magnitude of the normal vector
The direction cosines of the normal vector to the plane can be calculated using the formula:
\[ \text{Magnitude} = \sqrt{a^2 + b^2 + c^2} \]
Substituting the values:
\[ \text{Magnitude} = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 3: Calculate the direction cosines
The direction cosines \( l, m, n \) are given by:
\[ l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} \]
Substituting the values we have:
\[ l = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{-3}{\sqrt{14}} \]

### Step 4: Include the sign
Since direction cosines can be positive or negative, we include the plus-minus sign:
\[ l = \pm \frac{1}{\sqrt{14}}, \quad m = \pm \frac{2}{\sqrt{14}}, \quad n = \pm \frac{-3}{\sqrt{14}} \]

### Final Result
Thus, the direction cosines of the normal to the plane \( x + 2y - 3z + 4 = 0 \) are:
\[ \left( \pm \frac{1}{\sqrt{14}}, \pm \frac{2}{\sqrt{14}}, \pm \frac{-3}{\sqrt{14}} \right) \]