The vector equation of the line of intersection of the planes `vecr.(2hati+3hatk)=0` and `vecr.(3hati+2hatj+hatk)=0` is

Options

A. `vecr=lamda(hati+2hatj+hatk)`
B. `vecr=lamda(hati-2hatj+3hatk)`
C. `vecr=lamda(hati+2hatj-3hatk)`
D. none of these

Explanation

To find the vector equation of the line of intersection of the given planes, we can follow these steps:

### Step 1: Write the equations of the planes in Cartesian form
The given vector equations of the planes are:
1. \(\vec{r} \cdot (2\hat{i} + 3\hat{k}) = 0\)
2. \(\vec{r} \cdot (3\hat{i} + 2\hat{j} + \hat{k}) = 0\)

We can express \(\vec{r}\) as \(x\hat{i} + y\hat{j} + z\hat{k}\).

From the first plane:
\[ \vec{r} \cdot (2\hat{i} + 3\hat{k}) = 0 \implies 2x + 3z = 0 \quad \text{(Equation 1)} \]

From the second plane:
\[ \vec{r} \cdot (3\hat{i} + 2\hat{j} + \hat{k}) = 0 \implies 3x + 2y + z = 0 \quad \text{(Equation 2)} \]

### Step 2: Identify a point on the line of intersection
To find a point on the line of intersection, we can set \(z = 0\) in Equation 1:
\[ 2x + 3(0) = 0 \implies x = 0 \]
Substituting \(x = 0\) into Equation 2:
\[ 3(0) + 2y + 0 = 0 \implies 2y = 0 \implies y = 0 \]
Thus, the point of intersection is \((0, 0, 0)\).

### Step 3: Find the direction ratios of the line
The direction ratios of the line can be found using the cross product of the normals of the two planes. The normal vector of the first plane is \(\vec{n_1} = (2, 0, 3)\) and for the second plane, \(\vec{n_2} = (3, 2, 1)\).

We compute the cross product \(\vec{n_1} \times \vec{n_2}\):
\[ \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 3 & 2 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(0 \cdot 1 - 3 \cdot 2) - \hat{j}(2 \cdot 1 - 3 \cdot 3) + \hat{k}(2 \cdot 2 - 0 \cdot 3) \]
\[ = \hat{i}(0 - 6) - \hat{j}(2 - 9) + \hat{k}(4) \]
\[ = -6\hat{i} + 7\hat{j} + 4\hat{k} \]

Thus, the direction ratios of the line are \((-6, 7, 4)\).

### Step 4: Write the vector equation of the line
Using the point \((0, 0, 0)\) and the direction ratios \((-6, 7, 4)\), we can write the vector equation of the line as:
\[ \frac{x - 0}{-6} = \frac{y - 0}{7} = \frac{z - 0}{4} = \lambda \]
This can be simplified to:
\[ \frac{x}{-6} = \frac{y}{7} = \frac{z}{4} \]

### Final Answer
The vector equation of the line of intersection of the planes is:
\[ \frac{x}{-6} = \frac{y}{7} = \frac{z}{4} \]