The three planes `4y+6z=5,2x+3y+5z=5a n d6x+5y+9z=10` a. `` meet in a point b. have a line in common c. form a triangular prism d. none of these

Options

A. meet in a point
B. have a line in common
C. form a triangular prism
D. none of these

Explanation

To determine the relationship between the three given planes, we will analyze them step by step.

### Given Planes:
1. \( 4y + 6z = 5 \) (Plane 1)
2. \( 2x + 3y + 5z = 5 \) (Plane 2)
3. \( 6x + 5y - 9z = 10 \) (Plane 3)

### Step 1: Rewrite the equations in standard form
We can rewrite the equations in the standard form \( Ax + By + Cz = D \).

1. Plane 1: \( 0x + 4y + 6z = 5 \)
2. Plane 2: \( 2x + 3y + 5z = 5 \)
3. Plane 3: \( 6x + 5y - 9z = 10 \)

### Step 2: Formulate the coefficient matrix
The coefficient matrix \( A \) of the system of equations is:

\[ A = \begin{bmatrix} 0 & 4 & 6 \\ 2 & 3 & 5 \\ 6 & 5 & -9 \end{bmatrix} \]

### Step 3: Calculate the determinant of the coefficient matrix
To check if the planes are coplanar, we need to find the determinant of the matrix \( A \).

\[ \text{det}(A) = 0 \cdot \begin{vmatrix} 3 & 5 \\ 5 & -9 \end{vmatrix} - 4 \cdot \begin{vmatrix} 2 & 5 \\ 6 & -9 \end{vmatrix} + 6 \cdot \begin{vmatrix} 2 & 3 \\ 6 & 5 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 3 & 5 \\ 5 & -9 \end{vmatrix} = (3)(-9) - (5)(5) = -27 - 25 = -52 \)
2. \( \begin{vmatrix} 2 & 5 \\ 6 & -9 \end{vmatrix} = (2)(-9) - (5)(6) = -18 - 30 = -48 \)
3. \( \begin{vmatrix} 2 & 3 \\ 6 & 5 \end{vmatrix} = (2)(5) - (3)(6) = 10 - 18 = -8 \)

Now substituting back into the determinant:

\[ \text{det}(A) = 0 \cdot (-52) - 4 \cdot (-48) + 6 \cdot (-8) \]
\[ = 0 + 192 - 48 = 144 \]

### Step 4: Analyze the determinant
Since \( \text{det}(A) \neq 0 \), the planes are not coplanar. This means they do not intersect at a single point nor do they form a triangular prism. Instead, they intersect along a line.

### Conclusion
The three planes intersect along a line. Therefore, the correct option is:
**b. have a line in common.**

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