Find the equation of the plane containing the line. :
`(x -1)/(2) = (y -2)/(-1) = (z - 3)/(4)`
and perpendicular to the plane x + 2y + z - 2 = 0 .

Answer

To find the equation of the plane containing the line given by:

\[ \frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 3}{4} \]

and perpendicular to the plane defined by:

\[ x + 2y + z - 2 = 0, \]

we can follow these steps:

### Step 1: Identify the direction ratios of the line
The line can be expressed in parametric form. From the equation, we can identify the direction ratios of the line as \( \mathbf{L} = (2, -1, 4) \).

### Step 2: Identify the normal vector of the given plane
The normal vector of the plane \( x + 2y + z - 2 = 0 \) can be directly obtained from the coefficients of \( x, y, z \). Thus, the normal vector \( \mathbf{N_1} = (1, 2, 1) \).

### Step 3: Find the normal vector of the required plane
Since the required plane is perpendicular to the given plane, its normal vector \( \mathbf{N} \) must be perpendicular to \( \mathbf{N_1} \) and also to the direction ratios of the line \( \mathbf{L} \). We can find \( \mathbf{N} \) by taking the cross product of \( \mathbf{N_1} \) and \( \mathbf{L} \):

\[ \mathbf{N} = \mathbf{N_1} \times \mathbf{L} = (1, 2, 1) \times (2, -1, 4). \]

### Step 4: Calculate the cross product
Using the determinant method for the cross product:

\[ \mathbf{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 2 & -1 & 4 \end{vmatrix} \]

Calculating this determinant:

\[ \mathbf{N} = \mathbf{i} \begin{vmatrix} 2 & 1 \\ -1 & 4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 1 \\ -1 & 4 \end{vmatrix} = (2)(4) - (1)(-1) = 8 + 1 = 9 \)
2. \( \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix} = (1)(4) - (1)(2) = 4 - 2 = 2 \)
3. \( \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \)

Putting it all together:

\[ \mathbf{N} = 9\mathbf{i} - 2\mathbf{j} - 5\mathbf{k} = (9, -2, -5). \]

### Step 5: Use a point on the line
The line passes through the point \( (1, 2, 3) \). We can use this point to find the equation of the plane.

### Step 6: Write the equation of the plane
The equation of a plane can be expressed as:

\[ N_x(x - x_0) + N_y(y - y_0) + N_z(z - z_0) = 0, \]

where \( (x_0, y_0, z_0) \) is a point on the plane and \( (N_x, N_y, N_z) \) are the components of the normal vector.

Substituting the values:

\[ 9(x - 1) - 2(y - 2) - 5(z - 3) = 0. \]

### Step 7: Expand and simplify
Expanding this equation:

\[ 9x - 9 - 2y + 4 - 5z + 15 = 0, \]

which simplifies to:

\[ 9x - 2y - 5z + 10 = 0. \]

### Final Equation
Thus, the equation of the required plane is:

\[ 9x - 2y - 5z + 10 = 0. \]