The Cartesian equation of the plane `vecr=(1+lamda-mu)hati+(2-lamda+mu)hatj+(3-2lamda+2mu)hatk` is

Options

A. `2x+y=5`
B. `2x-y=5`
C. `2x+z=5`
D. `2x-z=5`

Explanation

To find the Cartesian equation of the plane given by the vector equation \(\vec{r} = (1 + \lambda - \mu) \hat{i} + (2 - \lambda + \mu) \hat{j} + (3 - 2\lambda + 2\mu) \hat{k}\), we can follow these steps:

### Step 1: Identify the point and direction vectors
The vector equation can be rewritten in the form:
\[ \vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c} \]
where \(\vec{a}\) is a position vector of a point on the plane, and \(\vec{b}\) and \(\vec{c}\) are direction vectors.

From the equation, we can extract:
- The point \(\vec{a} = (1, 2, 3)\)
- The direction vector associated with \(\lambda\) is \(\vec{b} = (1, -1, -2)\)
- The direction vector associated with \(\mu\) is \(\vec{c} = (-1, 1, 2)\)

### Step 2: Formulate the Cartesian equation
To find the Cartesian equation of the plane, we need to express the relationship between \(x\), \(y\), and \(z\). The general form of the Cartesian equation of a plane given by two direction vectors \(\vec{b}\) and \(\vec{c}\) is:
\[ \frac{x - x_0}{b_1} = \frac{y - y_0}{b_2} = \frac{z - z_0}{b_3} \]
where \((x_0, y_0, z_0)\) is a point on the plane and \((b_1, b_2, b_3)\) are the components of the direction vectors.

Using the point \((1, 2, 3)\) and the direction vector \(\vec{b} = (1, -1, -2)\), we can write:
\[ \frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 3}{-2} \]

### Step 3: Write the final Cartesian equation
Thus, the Cartesian equation of the plane can be expressed as:
\[ \frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 3}{-2} \]

### Final Answer
The Cartesian equation of the plane is:
\[ \frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 3}{-2} \]