The vector equation of the plane containing he line `vecr=(-2hati-3hatj+4hatk)+lamda(3hati-2hatj-hatk)` and the point `hati+2hatj+3hatk` is

Options

A. `vecr.(hati+3hatk)=10`
B. `vecr.(hati-3hatk)=10`
C. ` vecr.(3hati+hatk)=10`
D. none of these

Explanation

To find the vector equation of the plane containing the given line and point, we can follow these steps:

### Step 1: Identify Points and Direction Vector
The line is given by the equation:
\[ \vec{r} = (-2\hat{i} - 3\hat{j} + 4\hat{k}) + \lambda(3\hat{i} - 2\hat{j} - \hat{k}) \]
From this, we can identify:
- A point on the line, \( P = (-2, -3, 4) \)
- The direction vector of the line, \( \vec{b} = (3, -2, -1) \)

Additionally, we have another point:
\[ Q = (1, 2, 3) \]

### Step 2: Find the Vector \( \vec{PQ} \)
Next, we need to find the vector \( \vec{PQ} \) from point \( P \) to point \( Q \):
\[ \vec{PQ} = Q - P = (1 - (-2), 2 - (-3), 3 - 4) = (1 + 2, 2 + 3, 3 - 4) = (3, 5, -1) \]

### Step 3: Find the Normal Vector
To find the normal vector \( \vec{n} \) to the plane, we can take the cross product of the two vectors \( \vec{b} \) and \( \vec{PQ} \):
\[ \vec{n} = \vec{PQ} \times \vec{b} \]
Calculating the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & -1 \\ 3 & -2 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{n} = \hat{i}(5 \cdot (-1) - (-1) \cdot (-2)) - \hat{j}(3 \cdot (-1) - (-1) \cdot 3) + \hat{k}(3 \cdot (-2) - 5 \cdot 3) \]
\[ = \hat{i}(-5 - 2) - \hat{j}(-3 - 3) + \hat{k}(-6 - 15) \]
\[ = -7\hat{i} + 6\hat{j} - 21\hat{k} \]

### Step 4: Equation of the Plane
The equation of the plane can be written in the form:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
where \( \vec{a} \) is a point on the plane (we can use point \( P \)):
\[ \vec{a} = (-2, -3, 4) \]
Calculating \( \vec{a} \cdot \vec{n} \):
\[ \vec{a} \cdot \vec{n} = (-2)(-7) + (-3)(6) + (4)(-21) = 14 - 18 - 84 = -88 \]
Thus, the equation of the plane becomes:
\[ \vec{r} \cdot (-7\hat{i} + 6\hat{j} - 21\hat{k}) = -88 \]

### Step 5: Final Form
To express it in a standard form, we can rearrange:
\[ -7x + 6y - 21z = -88 \]
or equivalently:
\[ 7x - 6y + 21z = 88 \]

### Summary of the Vector Equation of the Plane
The vector equation of the plane is:
\[ \vec{r} \cdot (-7\hat{i} + 6\hat{j} - 21\hat{k}) = -88 \]