Distance between two non - intersecting planes `P_(1) and P_(2)` is 5 units, where `P_(1)` is `2x-3y+6z+26=0` and `P_(2)` is `4x+by+cz+d=0`. The point `A(-3, 0,-1)` lies between the planes `P_(1)` and `P_(2)`, then the value of `3b+4c-5d` is equal to

Options

A. 580
B. 120
C. `-18`
D. `-120`

Answer

To solve the problem, we need to find the value of \(3b + 4c - 5d\) given the planes \(P_1\) and \(P_2\) and the point \(A(-3, 0, -1)\).

### Step-by-Step Solution:

1. **Identify the equations of the planes**:
- The first plane \(P_1\) is given by the equation:
\[ 2x - 3y + 6z + 26 = 0 \]
- The second plane \(P_2\) is given by the equation:
\[ 4x + by + cz + d = 0 \]

2. **Determine the normal vectors**:
- The normal vector of \(P_1\) is \(\vec{n_1} = (2, -3, 6)\).
- The normal vector of \(P_2\) is \(\vec{n_2} = (4, b, c)\).

3. **Check if the planes are parallel**:
- For the planes to be parallel, the normal vectors must be proportional. Thus, we can write:
\[ \frac{2}{4} = \frac{-3}{b} = \frac{6}{c} \]
- From \(\frac{2}{4} = \frac{1}{2}\), we have:
\[ b = -6 \quad \text{(from } \frac{-3}{b} = \frac{1}{2} \text{)} \]
- From \(\frac{6}{c} = \frac{1}{2}\), we have:
\[ c = 12 \]

4. **Find the distance between the planes**:
- The distance \(d\) between two parallel planes \(Ax + By + Cz + D_1 = 0\) and \(Ax + By + Cz + D_2 = 0\) is given by:
\[ \text{Distance} = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]
- For our planes:
- \(D_1 = 26\) (from \(P_1\))
- \(D_2 = d\) (from \(P_2\))
- The coefficients are \(A = 4\), \(B = -6\), \(C = 12\).
- The distance is given as 5 units:
\[ \frac{|d - 26|}{\sqrt{4^2 + (-6)^2 + 12^2}} = 5 \]
- Calculate the denominator:
\[ \sqrt{16 + 36 + 144} = \sqrt{196} = 14 \]
- Thus, we have:
\[ \frac{|d - 26|}{14} = 5 \]
- This leads to:
\[ |d - 26| = 70 \]
- Therefore, we have two cases:
\[ d - 26 = 70 \quad \Rightarrow \quad d = 96 \]
\[ d - 26 = -70 \quad \Rightarrow \quad d = -44 \]

5. **Determine which \(d\) value to use**:
- Since point \(A(-3, 0, -1)\) lies between the planes, we need to check the signs of the equations when substituting \(A\).
- For \(d = 96\):
\[ 4(-3) - 6(0) + 12(-1) + 96 = -12 + 0 - 12 + 96 = 72 \quad (\text{positive}) \]
- For \(d = -44\):
\[ 4(-3) - 6(0) + 12(-1) - 44 = -12 + 0 - 12 - 44 = -68 \quad (\text{negative}) \]
- Since the point \(A\) is between the planes, we take \(d = -44\).

6. **Substitute values into \(3b + 4c - 5d\)**:
- We have \(b = -6\), \(c = 12\), and \(d = -44\):
\[ 3b + 4c - 5d = 3(-6) + 4(12) - 5(-44) \]
- Calculate:
\[ = -18 + 48 + 220 = 250 \]

### Final Answer:
The value of \(3b + 4c - 5d\) is \(250\).