Find the value of `'lambda'` if the following planes are perpendicular.
(i) `2x-4y+3z+1 = 0` and `x +2y+lambdaz = 3`
(ii) `vecr.(3hati-6hatj-2hatk) = 1` and
`vecr.(2hati+hatj-lambdahati) = 2`

Answer

To find the value of `lambda` for the given planes to be perpendicular, we will solve the problem step by step.

### Part (i)

1. **Identify the normal vectors from the plane equations.**
- The first plane is given by the equation:
\[ 2x - 4y + 3z + 1 = 0 \]
The normal vector \( \vec{n_1} \) can be extracted as:
\[ \vec{n_1} = \langle 2, -4, 3 \rangle \]

- The second plane is given by the equation:
\[ x + 2y + \lambda z = 3 \]
Rearranging it gives:
\[ x + 2y + \lambda z - 3 = 0 \]
The normal vector \( \vec{n_2} \) is:
\[ \vec{n_2} = \langle 1, 2, \lambda \rangle \]

2. **Set up the condition for perpendicularity.**
- For the planes to be perpendicular, their normal vectors must also be perpendicular. This means:
\[ \vec{n_1} \cdot \vec{n_2} = 0 \]

3. **Calculate the dot product.**
- The dot product is calculated as follows:
\[ \vec{n_1} \cdot \vec{n_2} = (2)(1) + (-4)(2) + (3)(\lambda) = 0 \]
- Simplifying this gives:
\[ 2 - 8 + 3\lambda = 0 \]
\[ 3\lambda - 6 = 0 \]

4. **Solve for `lambda`.**
- Rearranging gives:
\[ 3\lambda = 6 \implies \lambda = 2 \]

### Part (ii)

1. **Identify the normal vectors from the plane equations.**
- The first plane is given by:
\[ \vec{r} \cdot (3\hat{i} - 6\hat{j} - 2\hat{k}) = 1 \]
The normal vector \( \vec{n_1} \) is:
\[ \vec{n_1} = \langle 3, -6, -2 \rangle \]

- The second plane is given by:
\[ \vec{r} \cdot (2\hat{i} + \hat{j} - \lambda\hat{k}) = 2 \]
The normal vector \( \vec{n_2} \) is:
\[ \vec{n_2} = \langle 2, 1, -\lambda \rangle \]

2. **Set up the condition for perpendicularity.**
- Again, for the planes to be perpendicular:
\[ \vec{n_1} \cdot \vec{n_2} = 0 \]

3. **Calculate the dot product.**
- The dot product is:
\[ \vec{n_1} \cdot \vec{n_2} = (3)(2) + (-6)(1) + (-2)(-\lambda) = 0 \]
- Simplifying gives:
\[ 6 - 6 + 2\lambda = 0 \]
\[ 2\lambda = 0 \]

4. **Solve for `lambda`.**
- This gives:
\[ \lambda = 0 \]

### Final Answers
- For part (i), \( \lambda = 2 \)
- For part (ii), \( \lambda = 0 \)