Find the value of 'k' for which the plane :
3x - 6y - 2z = 7 and 2x + y - kz = 3
are perpendicular to each other.

Answer

To find the value of \( k \) for which the planes \( 3x - 6y - 2z = 7 \) and \( 2x + y - kz = 3 \) are perpendicular to each other, we follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of the first plane \( 3x - 6y - 2z = 7 \) can be derived from the coefficients of \( x \), \( y \), and \( z \):
\[ \mathbf{n_1} = \langle 3, -6, -2 \rangle \]

The normal vector of the second plane \( 2x + y - kz = 3 \) is:
\[ \mathbf{n_2} = \langle 2, 1, -k \rangle \]

### Step 2: Use the condition for perpendicularity
Two planes are perpendicular if the dot product of their normal vectors is zero:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 0 \]

Calculating the dot product:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (3)(2) + (-6)(1) + (-2)(-k) \]
\[ = 6 - 6 + 2k \]
\[ = 2k \]

### Step 3: Set the dot product to zero
To find the value of \( k \), we set the dot product equal to zero:
\[ 2k = 0 \]

### Step 4: Solve for \( k \)
Dividing both sides by 2:
\[ k = 0 \]

Thus, the value of \( k \) for which the planes are perpendicular is:
\[ \boxed{0} \]

---