If `d_(1),d_(2),d_(3)` denote the distances of the plane `2x-3y+4z=0` from the planes `2x-3y+4z+6=0` `4x-6y+8z+3=0` and `2x-3y+4z-6=0` respectively, then

Options

A. `d_(1)+8d_(2)+d_(3)=0`
B. `d_(1)+16d_(2)=0`
C. `8d_(2)=d_(1)`
D. `d_(1)-2d_(2)+3d_(3)=sqrt(29)`

Explanation

To find the distances \( d_1, d_2, d_3 \) from the plane \( 2x - 3y + 4z = 0 \) to the planes \( 2x - 3y + 4z + 6 = 0 \), \( 4x - 6y + 8z + 3 = 0 \), and \( 2x - 3y + 4z - 6 = 0 \), we will use the formula for the distance between two parallel planes.

### Step 1: Identify the planes
The given planes are:
1. Plane \( P_1: 2x - 3y + 4z = 0 \)
2. Plane \( P_2: 2x - 3y + 4z + 6 = 0 \)
3. Plane \( P_3: 4x - 6y + 8z + 3 = 0 \)
4. Plane \( P_4: 2x - 3y + 4z - 6 = 0 \)

### Step 2: Calculate \( d_1 \) (distance from \( P_1 \) to \( P_2 \))
Using the formula for the distance between two parallel planes:
\[ d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \]
where \( a, b, c \) are the coefficients of \( x, y, z \) respectively.

For planes \( P_1 \) and \( P_2 \):
- \( d_1 = 0 \) (from \( P_1 \))
- \( d_2 = 6 \) (from \( P_2 \))
- Coefficients: \( a = 2, b = -3, c = 4 \)

Calculating \( d_1 \):
\[ d_1 = \frac{|6 - 0|}{\sqrt{2^2 + (-3)^2 + 4^2}} = \frac{6}{\sqrt{4 + 9 + 16}} = \frac{6}{\sqrt{29}} \]

### Step 3: Calculate \( d_2 \) (distance from \( P_1 \) to \( P_3 \))
First, rewrite \( P_3 \) in the same form as \( P_1 \):
\[ 4x - 6y + 8z + 3 = 0 \implies 2x - 3y + 4z + \frac{3}{2} = 0 \]
Now, using the formula:
- \( d_1 = 0 \)
- \( d_2 = \frac{3}{2} \)

Calculating \( d_2 \):
\[ d_2 = \frac{\left| \frac{3}{2} - 0 \right|}{\sqrt{2^2 + (-3)^2 + 4^2}} = \frac{\frac{3}{2}}{\sqrt{29}} = \frac{3}{2\sqrt{29}} \]

### Step 4: Calculate \( d_3 \) (distance from \( P_1 \) to \( P_4 \))
For planes \( P_1 \) and \( P_4 \):
- \( d_1 = 0 \)
- \( d_2 = -6 \) (from \( P_4 \))

Calculating \( d_3 \):
\[ d_3 = \frac{|0 - (-6)|}{\sqrt{2^2 + (-3)^2 + 4^2}} = \frac{6}{\sqrt{29}} \]

### Step 5: Establish the relationship between \( d_1, d_2, d_3 \)
We have:
- \( d_1 = \frac{6}{\sqrt{29}} \)
- \( d_2 = \frac{3}{2\sqrt{29}} \)
- \( d_3 = \frac{8}{\sqrt{29}} \)

Now, let's check the relationship:
\[ d_1 = 8 \cdot d_2 \implies \frac{6}{\sqrt{29}} = 8 \cdot \frac{3}{2\sqrt{29}} \]
Calculating \( 8 \cdot d_2 \):
\[ 8 \cdot \frac{3}{2\sqrt{29}} = \frac{24}{2\sqrt{29}} = \frac{12}{\sqrt{29}} \]
Thus, the relationship \( d_1 = 8d_2 \) holds.

### Final Result
The relationship between the distances is:
\[ d_1 = 8d_2 \]