The plane containing the line `(x-1)/(1)=(y-2)/(2)=(z-3)/(3)` and parallel to the line `(x)/(1)=(y)/(1)=(z)/(4)` passes through the point

Options

A. `(1,-2,5)`
B. `(1,0,5)`
C. `(0,3,-5)`
D. `(-1,-3,0)`

Answer

To solve the problem, we need to find the equation of a plane that contains a given line and is parallel to another line, while also passing through a specific point.

### Step-by-Step Solution:

1. **Identify the Given Lines and Point:**
- The first line is given by the equation \((x-1)/(1)=(y-2)/(2)=(z-3)/(3)\). This can be represented in parametric form as:
\[ x = 1 + t, \quad y = 2 + 2t, \quad z = 3 + 3t \]
where \(t\) is a parameter.
- The second line is given by \((x)/(1)=(y)/(1)=(z)/(4)\), which can be represented as:
\[ x = s, \quad y = s, \quad z = 4s \]
where \(s\) is another parameter.
- The point through which the plane passes is \(P(1, 2, 3)\).

2. **Determine Direction Ratios:**
- The direction ratios of the first line (let's call it Line 1) are \((1, 2, 3)\).
- The direction ratios of the second line (Line 2) are \((1, 1, 4)\).

3. **Find the Normal Vector of the Plane:**
- Since the plane contains Line 1 and is parallel to Line 2, we can find a normal vector to the plane by taking the cross product of the direction ratios of the two lines.
- Let \(\mathbf{a} = (1, 2, 3)\) and \(\mathbf{b} = (1, 1, 4)\).
- The cross product \(\mathbf{n} = \mathbf{a} \times \mathbf{b}\) is calculated as follows:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 1 & 1 & 4 \end{vmatrix} \]
- Calculating the determinant gives:
\[ \mathbf{n} = \mathbf{i}(2 \cdot 4 - 3 \cdot 1) - \mathbf{j}(1 \cdot 4 - 3 \cdot 1) + \mathbf{k}(1 \cdot 1 - 2 \cdot 1) \]
\[ = \mathbf{i}(8 - 3) - \mathbf{j}(4 - 3) + \mathbf{k}(1 - 2) \]
\[ = 5\mathbf{i} - 1\mathbf{j} - 1\mathbf{k} \]
- Thus, the normal vector \(\mathbf{n} = (5, -1, -1)\).

4. **Equation of the Plane:**
- The equation of a plane can be expressed in the form:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
- Substituting the normal vector \((5, -1, -1)\) and the point \(P(1, 2, 3)\):
\[ 5(x - 1) - 1(y - 2) - 1(z - 3) = 0 \]
- Expanding this gives:
\[ 5x - 5 - y + 2 - z + 3 = 0 \]
\[ 5x - y - z = 0 \]

5. **Final Equation of the Plane:**
- The equation of the plane is:
\[ 5x - y - z = 0 \]