If `vec(a)=-2hat(i)+3hat(j)+5hat(k), vec(b)=hat(i)+2hat(j)+3hat(k)` and `vec(c )=7 hat(i)-hat(k)` are position vectors of three points A, B, C respectively, prove that A, B, C are collinear.

Answer

To prove that the points A, B, and C represented by the position vectors \(\vec{a} = -2\hat{i} + 3\hat{j} + 5\hat{k}\), \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\), and \(\vec{c} = 7\hat{i} - \hat{k}\) are collinear, we can follow these steps:

### Step 1: Find the vector \(\vec{AB}\)
The vector \(\vec{AB}\) can be found using the formula:
\[ \vec{AB} = \vec{b} - \vec{a} \]
Substituting the values of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{AB} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (-2\hat{i} + 3\hat{j} + 5\hat{k}) \]
Calculating this gives:
\[ \vec{AB} = \hat{i} + 2\hat{j} + 3\hat{k} + 2\hat{i} - 3\hat{j} - 5\hat{k} \]
\[ \vec{AB} = (1 + 2)\hat{i} + (2 - 3)\hat{j} + (3 - 5)\hat{k} \]
\[ \vec{AB} = 3\hat{i} - \hat{j} - 2\hat{k} \]

### Step 2: Find the vector \(\vec{BC}\)
Next, we find the vector \(\vec{BC}\):
\[ \vec{BC} = \vec{c} - \vec{b} \]
Substituting the values of \(\vec{b}\) and \(\vec{c}\):
\[ \vec{BC} = (7\hat{i} - \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) \]
Calculating this gives:
\[ \vec{BC} = 7\hat{i} - \hat{k} - \hat{i} - 2\hat{j} - 3\hat{k} \]
\[ \vec{BC} = (7 - 1)\hat{i} - 2\hat{j} + (-1 - 3)\hat{k} \]
\[ \vec{BC} = 6\hat{i} - 2\hat{j} - 4\hat{k} \]

### Step 3: Check for collinearity
To check if the vectors \(\vec{AB}\) and \(\vec{BC}\) are collinear, we need to see if \(\vec{BC}\) is a scalar multiple of \(\vec{AB}\). We can express this as:
\[ \vec{BC} = k \cdot \vec{AB} \]
for some scalar \(k\).

From our calculations:
\[ \vec{AB} = 3\hat{i} - \hat{j} - 2\hat{k} \]
\[ \vec{BC} = 6\hat{i} - 2\hat{j} - 4\hat{k} \]

Now, we can find \(k\) by comparing components:
- For \(\hat{i}\): \(6 = 2 \cdot 3\) implies \(k = 2\)
- For \(\hat{j}\): \(-2 = 2 \cdot -1\) confirms \(k = 2\)
- For \(\hat{k}\): \(-4 = 2 \cdot -2\) confirms \(k = 2\)

Since all components yield the same scalar \(k\), we conclude that:
\[ \vec{BC} = 2 \cdot \vec{AB} \]

### Conclusion
Since \(\vec{BC}\) is a scalar multiple of \(\vec{AB}\), the points A, B, and C are collinear.

---