Find the value of `m` for which the straight line `3x-2y+z+3=0=4x-3y+4z+1` is parallel to the plane `2x-y+m z-2=0.`

Options

A. -2
B. 8
C. -18
D. 11

Explanation

To find the value of \( m \) for which the straight line defined by the equations \( 3x - 2y + z + 3 = 0 \) and \( 4x - 3y + 4z + 1 = 0 \) is parallel to the plane \( 2x - y + mz - 2 = 0 \), we can follow these steps:

### Step 1: Identify the direction ratios of the line

The given equations of the line can be rewritten in the form of two planes. The direction ratios of the line can be found using the cross product of the normals of these two planes.

The normal vector of the first plane \( 3x - 2y + z + 3 = 0 \) is \( \mathbf{n_1} = (3, -2, 1) \).

The normal vector of the second plane \( 4x - 3y + 4z + 1 = 0 \) is \( \mathbf{n_2} = (4, -3, 4) \).

### Step 2: Compute the cross product

To find the direction ratios of the line, we compute the cross product \( \mathbf{n_1} \times \mathbf{n_2} \):

\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 1 \\ 4 & -3 & 4 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{n_1} \times \mathbf{n_2} = \mathbf{i}((-2)(4) - (1)(-3)) - \mathbf{j}((3)(4) - (1)(4)) + \mathbf{k}((3)(-3) - (-2)(4)) \]

Calculating each component:

- For \( \mathbf{i} \): \( -8 + 3 = -5 \)
- For \( \mathbf{j} \): \( 12 - 4 = 8 \) (note the negative sign in front)
- For \( \mathbf{k} \): \( -9 + 8 = -1 \)

Thus, the direction ratios of the line are \( (-5, -8, -1) \).

### Step 3: Find the normal vector of the plane

The normal vector of the plane \( 2x - y + mz - 2 = 0 \) is \( \mathbf{n_3} = (2, -1, m) \).

### Step 4: Set up the condition for parallelism

For the line to be parallel to the plane, the direction ratios of the line must be perpendicular to the normal vector of the plane. This means their dot product must be zero:

\[ (-5, -8, -1) \cdot (2, -1, m) = 0 \]

Calculating the dot product:

\[ -5 \cdot 2 + (-8)(-1) + (-1)(m) = 0 \]

This simplifies to:

\[ -10 + 8 - m = 0 \]

### Step 5: Solve for \( m \)

Rearranging gives:

\[ -2 - m = 0 \implies m = -2 \]

### Final Answer

Thus, the value of \( m \) for which the straight line is parallel to the plane is:

\[ \boxed{-2} \]