The position vectors of the vertices `A, B and C` of triangle are `hati+hatj, hatj+hatk and hati+hatk`, respectively. Find the unit vectors `hatr` lying in the plane of `ABC` and perpendicular to `IA`, where I is the incentre of the triangle.

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Identify the position vectors of the vertices A, B, and C.

The position vectors are given as:
- \( \vec{A} = \hat{i} + \hat{j} \)
- \( \vec{B} = \hat{j} + \hat{k} \)
- \( \vec{C} = \hat{i} + \hat{k} \)

### Step 2: Find the vectors AB, BC, and AC.

1. **Vector AB**:
\[ \vec{AB} = \vec{B} - \vec{A} = (\hat{j} + \hat{k}) - (\hat{i} + \hat{j}) = -\hat{i} + 0\hat{j} + \hat{k} = -\hat{i} + \hat{k} \]

2. **Vector BC**:
\[ \vec{BC} = \vec{C} - \vec{B} = (\hat{i} + \hat{k}) - (\hat{j} + \hat{k}) = \hat{i} - \hat{j} + 0\hat{k} = \hat{i} - \hat{j} \]

3. **Vector AC**:
\[ \vec{AC} = \vec{C} - \vec{A} = (\hat{i} + \hat{k}) - (\hat{i} + \hat{j}) = 0\hat{i} - \hat{j} + \hat{k} = -\hat{j} + \hat{k} \]

### Step 3: Find the magnitudes of the vectors.

1. **Magnitude of AB**:
\[ |\vec{AB}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \]

2. **Magnitude of BC**:
\[ |\vec{BC}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \]

3. **Magnitude of AC**:
\[ |\vec{AC}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

### Step 4: Conclude that triangle ABC is equilateral.

Since all sides have the same length \( \sqrt{2} \), triangle ABC is equilateral.

### Step 5: Find the incenter I of triangle ABC.

In an equilateral triangle, the incenter coincides with the centroid and circumcenter. Therefore, the incenter \( I \) is at the average of the vertices:
\[ \vec{I} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = \frac{(\hat{i} + \hat{j}) + (\hat{j} + \hat{k}) + (\hat{i} + \hat{k})}{3} = \frac{2\hat{i} + 2\hat{j} + 2\hat{k}}{3} = \frac{2}{3}(\hat{i} + \hat{j} + \hat{k}) \]

### Step 6: Find the unit vector \( \hat{r} \) lying in the plane of ABC and perpendicular to \( \vec{IA} \).

1. **Vector IA**:
\[ \vec{IA} = \vec{A} - \vec{I} = \left(\hat{i} + \hat{j}\right) - \frac{2}{3}(\hat{i} + \hat{j} + \hat{k}) = \left(\hat{i} + \hat{j}\right) - \left(\frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\right) \]
\[ = \left(\hat{i} - \frac{2}{3}\hat{i}\right) + \left(\hat{j} - \frac{2}{3}\hat{j}\right) - \frac{2}{3}\hat{k} = \frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} \]

2. **Finding a vector perpendicular to IA**:
A vector perpendicular to \( \vec{IA} \) can be found using the cross product with another vector in the plane, such as \( \vec{BC} \).

3. **Unit vector \( \hat{r} \)**:
We can take \( \vec{BC} \) as \( \hat{i} - \hat{j} \) and find the unit vector:
\[ \hat{r} = \frac{\vec{BC}}{|\vec{BC}|} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} \]

### Final Answer:
The unit vector \( \hat{r} \) lying in the plane of triangle ABC and perpendicular to \( \vec{IA} \) is:
\[ \hat{r} = \frac{1}{\sqrt{2}}(\hat{i} - \hat{j}) \]