Suppose `AB=i+2j+4k" and " AC=5i+j+2k` are two sides of a triangle ABC whose centroid is G, then `abs(AG)=` __________

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the vectors
We are given:
- Vector \( \vec{AB} = \mathbf{i} + 2\mathbf{j} + 4\mathbf{k} \)
- Vector \( \vec{AC} = 5\mathbf{i} + \mathbf{j} + 2\mathbf{k} \)

### Step 2: Find the centroid \( G \) of triangle \( ABC \)
The formula for the centroid \( G \) of triangle \( ABC \) is given by:
\[ \vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]
In terms of vectors \( \vec{AB} \) and \( \vec{AC} \), we can express \( \vec{G} \) as:
\[ \vec{G} = \frac{\vec{AB} + \vec{AC}}{3} \]

### Step 3: Substitute the vectors into the centroid formula
Substituting the given vectors:
\[ \vec{G} = \frac{(\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) + (5\mathbf{i} + \mathbf{j} + 2\mathbf{k})}{3} \]

### Step 4: Combine the vectors
Now, combine the components:
\[ \vec{G} = \frac{(1 + 5)\mathbf{i} + (2 + 1)\mathbf{j} + (4 + 2)\mathbf{k}}{3} \]
This simplifies to:
\[ \vec{G} = \frac{6\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}}{3} \]

### Step 5: Simplify the centroid vector
Dividing each component by 3 gives:
\[ \vec{G} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k} \]

### Step 6: Find vector \( \vec{AG} \)
To find \( \vec{AG} \), we need to express it in terms of \( \vec{A} \) and \( \vec{G} \). Since \( \vec{A} \) is the origin in this context, we have:
\[ \vec{AG} = \vec{G} - \vec{A} = \vec{G} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k} \]

### Step 7: Calculate the modulus of \( \vec{AG} \)
The modulus of a vector \( \vec{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is given by:
\[ |\vec{v}| = \sqrt{a^2 + b^2 + c^2} \]
For \( \vec{AG} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k} \):
\[ |\vec{AG}| = \sqrt{(2)^2 + (1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Final Answer
Thus, the modulus of \( \vec{AG} \) is:
\[ \boxed{3} \]