Find the vector of length 3 unit which is perpendicular to ` hat i+ hat j+ hat k` and lies in the plane of ` hat i+ hat j+ hat ka n d2 hat k-3 hat j` .

Answer

To solve the problem, we need to find a vector of length 3 units that is perpendicular to the vector \( \hat{i} + \hat{j} + \hat{k} \) and lies in the plane defined by the vectors \( \hat{i} + \hat{j} + \hat{k} \) and \( 2\hat{k} - 3\hat{j} \).

### Step-by-Step Solution:

1. **Assume a General Vector**:
Let the vector we are looking for be \( \vec{A} = a\hat{i} + b\hat{j} + c\hat{k} \).

2. **Determine the Plane**:
The vector \( \vec{A} \) lies in the plane defined by the vectors \( \vec{B} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{C} = 2\hat{k} - 3\hat{j} \).
We can express \( \vec{A} \) as a linear combination of \( \vec{B} \) and \( \vec{C} \):
\[ \vec{A} = m(\hat{i} + \hat{j} + \hat{k}) + n(2\hat{k} - 3\hat{j}) \]

3. **Expand the Expression**:
Expanding the right-hand side:
\[ \vec{A} = m\hat{i} + m\hat{j} + m\hat{k} + 2n\hat{k} - 3n\hat{j} \]
Combining like terms:
\[ \vec{A} = m\hat{i} + (m - 3n)\hat{j} + (m + 2n)\hat{k} \]
Thus, we can identify:
- \( a = m \)
- \( b = m - 3n \)
- \( c = m + 2n \)

4. **Perpendicular Condition**:
The vector \( \vec{A} \) is perpendicular to \( \hat{i} + \hat{j} + \hat{k} \). Therefore, we have:
\[ a + b + c = 0 \]
Substituting the values of \( a, b, c \):
\[ m + (m - 3n) + (m + 2n) = 0 \]
Simplifying this gives:
\[ 3m - n = 0 \implies n = 3m \]

5. **Substituting for \( n \)**:
Substitute \( n = 3m \) back into the expressions for \( b \) and \( c \):
- For \( b \):
\[ b = m - 3(3m) = m - 9m = -8m \]
- For \( c \):
\[ c = m + 2(3m) = m + 6m = 7m \]

6. **Expressing \( \vec{A} \)**:
Now we have:
\[ \vec{A} = m\hat{i} - 8m\hat{j} + 7m\hat{k} \]
Factoring out \( m \):
\[ \vec{A} = m(\hat{i} - 8\hat{j} + 7\hat{k}) \]

7. **Finding the Magnitude**:
The magnitude of \( \vec{A} \) is given by:
\[ |\vec{A}| = m \sqrt{1^2 + (-8)^2 + 7^2} = m \sqrt{1 + 64 + 49} = m \sqrt{114} \]

8. **Setting the Magnitude to 3**:
We want the length of \( \vec{A} \) to be 3:
\[ m \sqrt{114} = 3 \implies m = \frac{3}{\sqrt{114}} = \frac{3\sqrt{114}}{114} = \frac{3}{\sqrt{114}} \]

9. **Final Vector**:
Substituting \( m \) back into \( \vec{A} \):
\[ \vec{A} = \frac{3}{\sqrt{114}}(\hat{i} - 8\hat{j} + 7\hat{k}) \]

### Conclusion:
The required vector of length 3 units that is perpendicular to \( \hat{i} + \hat{j} + \hat{k} \) and lies in the specified plane is:
\[ \vec{A} = \frac{3}{\sqrt{114}}(\hat{i} - 8\hat{j} + 7\hat{k}) \]