Find the perpendicular distance from the point P(2, 3, -4) to the plane `x+ 2y + 3z = 13.`

Answer

To find the perpendicular distance from the point \( P(2, 3, -4) \) to the plane given by the equation \( x + 2y + 3z = 13 \), we can use the formula for the distance from a point to a plane.

### Step-by-Step Solution:

1. **Identify the coefficients from the plane equation**:
The equation of the plane can be rewritten in the form \( ax + by + cz + d = 0 \).
Here, we have:
\[ x + 2y + 3z - 13 = 0 \]
Thus, \( a = 1 \), \( b = 2 \), \( c = 3 \), and \( d = -13 \).

2. **Identify the coordinates of the point**:
The coordinates of the point \( P \) are \( (x_1, y_1, z_1) = (2, 3, -4) \).

3. **Use the distance formula**:
The formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( ax + by + cz + d = 0 \) is given by:
\[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

4. **Substitute the values into the formula**:
We substitute \( a, b, c, d, x_1, y_1, z_1 \) into the formula:
\[ D = \frac{|1 \cdot 2 + 2 \cdot 3 + 3 \cdot (-4) - 13|}{\sqrt{1^2 + 2^2 + 3^2}} \]

5. **Calculate the numerator**:
Calculate \( ax_1 + by_1 + cz_1 + d \):
\[ = 2 + 6 - 12 - 13 = -17 \]
Thus, the absolute value is:
\[ |-17| = 17 \]

6. **Calculate the denominator**:
Calculate \( \sqrt{a^2 + b^2 + c^2} \):
\[ = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

7. **Final calculation of distance**:
Substitute the values back into the distance formula:
\[ D = \frac{17}{\sqrt{14}} \]

### Conclusion:
The perpendicular distance from the point \( P(2, 3, -4) \) to the plane \( x + 2y + 3z = 13 \) is:
\[ D = \frac{17}{\sqrt{14}} \text{ units} \]