The nature of the intersection of the set of planes:
`2x-4y+2z=5,5x-y-z=8` and `x+y-z=7` is / are that

Options

A. they form a triangluar prism
B. each of the plane is parallel to the line `(x)/(1)=(y)/(2)=(z)/(3)`
C. they intersect at a unique point
D. they pass through the line `(x)/(1)=(y)/(2)=(z)/(3)`.

Answer

To determine the nature of the intersection of the given set of planes, we need to analyze the equations of the planes and their normal vectors.

### Given Planes:
1. \( P_1: 2x - 4y + 2z = 5 \)
2. \( P_2: 5x - y - z = 8 \)
3. \( P_3: x + y - z = 7 \)

### Step 1: Identify the Normal Vectors
The normal vector of a plane given by the equation \( ax + by + cz = d \) is \( \vec{n} = (a, b, c) \).

- For \( P_1 \): \( \vec{n_1} = (2, -4, 2) \)
- For \( P_2 \): \( \vec{n_2} = (5, -1, -1) \)
- For \( P_3 \): \( \vec{n_3} = (1, 1, -1) \)

### Step 2: Form the Coefficient Matrix
We form the coefficient matrix using the normal vectors:

\[ \Delta = \begin{vmatrix} 2 & -4 & 2 \\ 5 & -1 & -1 \\ 1 & 1 & -1 \end{vmatrix} \]

### Step 3: Calculate the Determinant
To find the determinant \( \Delta \):

\[ \Delta = 2 \begin{vmatrix} -1 & -1 \\ 1 & -1 \end{vmatrix} - (-4) \begin{vmatrix} 5 & -1 \\ 1 & -1 \end{vmatrix} + 2 \begin{vmatrix} 5 & -1 \\ 1 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} -1 & -1 \\ 1 & -1 \end{vmatrix} = (-1)(-1) - (-1)(1) = 1 + 1 = 2 \)

2. \( \begin{vmatrix} 5 & -1 \\ 1 & -1 \end{vmatrix} = (5)(-1) - (-1)(1) = -5 + 1 = -4 \)

3. \( \begin{vmatrix} 5 & -1 \\ 1 & 1 \end{vmatrix} = (5)(1) - (-1)(1) = 5 + 1 = 6 \)

Now substituting back into the determinant calculation:

\[ \Delta = 2(2) + 4(-4) + 2(6) = 4 - 16 + 12 = 0 \]

### Step 4: Analyze the Result
Since \( \Delta = 0 \), this indicates that the planes do not intersect at a unique point. Instead, they either intersect along a line or form a triangular prism.

### Step 5: Check for Parallelism
To check if the planes are parallel to a specific line, we need to find the direction ratios of the line. Let's assume the line has direction ratios \( (1, 2, 3) \).

The direction vector \( \vec{b} = (1, 2, 3) \).

Now, we check if \( \vec{b} \) is perpendicular to each of the normal vectors:

1. \( \vec{b} \cdot \vec{n_1} = 1(2) + 2(-4) + 3(2) = 2 - 8 + 6 = 0 \)
2. \( \vec{b} \cdot \vec{n_2} = 1(5) + 2(-1) + 3(-1) = 5 - 2 - 3 = 0 \)
3. \( \vec{b} \cdot \vec{n_3} = 1(1) + 2(1) + 3(-1) = 1 + 2 - 3 = 0 \)

Since all dot products are zero, the line is parallel to all three planes.

### Conclusion
The nature of the intersection of the given planes is:
1. They form a triangular prism.
2. Each of the planes is parallel to the line.