If the equation of the plane passing through the point `(1,1,2)` and perpendicular to the line `x-3y+2z-1=0=4x-y+z` is `Ax+By+Cz=1`,then `140(C-B+A)` is equal to

Answer

To solve the problem, we need to find the equation of the plane that passes through the point \( (1, 1, 2) \) and is perpendicular to the line defined by the equations \( x - 3y + 2z - 1 = 0 \) and \( 4x - y + z = 0 \).

### Step 1: Find the direction ratios of the lines
The first step is to identify the direction ratios of the lines given by the equations.

1. The first line \( x - 3y + 2z - 1 = 0 \) can be represented in vector form. The coefficients of \( x, y, z \) give us the direction ratios:
- For \( x - 3y + 2z = 1 \), the direction ratios are \( (1, -3, 2) \).

2. The second line \( 4x - y + z = 0 \) also gives us direction ratios:
- For \( 4x - y + z = 0 \), the direction ratios are \( (4, -1, 1) \).

### Step 2: Find the normal to the plane
To find the normal to the plane, we take the cross product of the direction ratios of the two lines.

Let:
- \( \mathbf{n_1} = (1, -3, 2) \)
- \( \mathbf{n_2} = (4, -1, 1) \)

The cross product \( \mathbf{n_1} \times \mathbf{n_2} \) is calculated as follows:

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

Calculating the determinant:

\[ = \mathbf{i} \left( (-3)(1) - (2)(-1) \right) - \mathbf{j} \left( (1)(1) - (2)(4) \right) + \mathbf{k} \left( (1)(-1) - (-3)(4) \right) \]

\[ = \mathbf{i} (-3 + 2) - \mathbf{j} (1 - 8) + \mathbf{k} (-1 + 12) \]

\[ = \mathbf{i} (-1) + \mathbf{j} (7) + \mathbf{k} (11) \]

Thus, the normal vector to the plane is \( (-1, 7, 11) \).

### Step 3: Write the equation of the plane
The equation of a plane can be expressed as:

\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]

Where \( (x_0, y_0, z_0) \) is a point on the plane and \( (A, B, C) \) are the components of the normal vector.

Substituting the normal vector \( (-1, 7, 11) \) and the point \( (1, 1, 2) \):

\[ -1(x - 1) + 7(y - 1) + 11(z - 2) = 0 \]

Expanding this gives:

\[ -x + 1 + 7y - 7 + 11z - 22 = 0 \]

Combining like terms:

\[ -x + 7y + 11z - 28 = 0 \]

Rearranging gives:

\[ -x + 7y + 11z = 28 \]

### Step 4: Convert to the required form
We want the equation in the form \( Ax + By + Cz = 1 \). Dividing the entire equation by 28:

\[ -\frac{1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1 \]

Thus, we have:

\[ A = -\frac{1}{28}, \quad B = \frac{7}{28}, \quad C = \frac{11}{28} \]

### Step 5: Calculate \( 140(C - B + A) \)
Now, we need to calculate \( 140(C - B + A) \):

\[ C - B + A = \frac{11}{28} - \frac{7}{28} - \frac{1}{28} = \frac{11 - 7 - 1}{28} = \frac{3}{28} \]

Now multiplying by 140:

\[ 140 \left(\frac{3}{28}\right) = 5 \cdot 3 = 15 \]

### Final Answer
Thus, the value of \( 140(C - B + A) \) is \( \boxed{15} \).