Show that the points `A(-2 hat i+3 hat j+5 hat k), B( hat i+2 hat j+3 hat k)`and `C(7 hat i-3 hat k)`are collinear.

Explanation

To show that the points \( A(-2 \hat{i} + 3 \hat{j} + 5 \hat{k}), B(\hat{i} + 2 \hat{j} + 3 \hat{k}), C(7 \hat{i} - 3 \hat{k}) \) are collinear, we can use the determinant method. The points are collinear if the volume of the parallelepiped formed by the vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{BC} \) is zero, which can be determined using the determinant of a matrix formed by these vectors.

### Step 1: Find the position vectors of points A, B, and C.
- Position vector of \( A \): \( \vec{A} = -2 \hat{i} + 3 \hat{j} + 5 \hat{k} \)
- Position vector of \( B \): \( \vec{B} = \hat{i} + 2 \hat{j} + 3 \hat{k} \)
- Position vector of \( C \): \( \vec{C} = 7 \hat{i} + 0 \hat{j} - 3 \hat{k} \)

### Step 2: Find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).
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