If the planes `vecr.(2hati-hatj+2hatk)=4` and `vecr.(3hati+2hatj+lamda hatk)=3` are perpendicular then `lamda=`

Options

A. `2`
B. `-2`
C. `3`
D. `-3`

Explanation

To solve the problem, we need to find the value of \(\lambda\) such that the two given planes are perpendicular. The equations of the planes are:

1. \(\vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 4\)
2. \(\vec{r} \cdot (3\hat{i} + 2\hat{j} + \lambda \hat{k}) = 3\)

### Step 1: Identify the normals of the planes
The normal vector of the first plane, \( \vec{n_1} \), is given by the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) in the first equation:
\[ \vec{n_1} = 2\hat{i} - \hat{j} + 2\hat{k} \]

The normal vector of the second plane, \( \vec{n_2} \), is given by the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) in the second equation:
\[ \vec{n_2} = 3\hat{i} + 2\hat{j} + \lambda \hat{k} \]

### Step 2: Use the condition for perpendicularity
For the two planes to be perpendicular, their normal vectors must also be perpendicular. This means that their dot product must equal zero:
\[ \vec{n_1} \cdot \vec{n_2} = 0 \]

### Step 3: Calculate the dot product
Now, we calculate the dot product:
\[ \vec{n_1} \cdot \vec{n_2} = (2\hat{i} - \hat{j} + 2\hat{k}) \cdot (3\hat{i} + 2\hat{j} + \lambda \hat{k}) \]

Calculating the dot product:
\[ = 2 \cdot 3 + (-1) \cdot 2 + 2 \cdot \lambda \]
\[ = 6 - 2 + 2\lambda \]
\[ = 4 + 2\lambda \]

### Step 4: Set the dot product to zero
Setting the dot product equal to zero gives us:
\[ 4 + 2\lambda = 0 \]

### Step 5: Solve for \(\lambda\)
Now, we can solve for \(\lambda\):
\[ 2\lambda = -4 \]
\[ \lambda = -2 \]

### Conclusion
Thus, the value of \(\lambda\) is:
\[ \lambda = -2 \]