Show that vector `hati + 3hatj + hatk, 2 hati - hatj - hatk, 7 hatj + 3 hatk` are parallel to same plane.

Answer

To show that the vectors \(\hat{i} + 3\hat{j} + \hat{k}\), \(2\hat{i} - \hat{j} - \hat{k}\), and \(7\hat{j} + 3\hat{k}\) are parallel to the same plane, we need to demonstrate that these vectors are coplanar. A set of three vectors is coplanar if their scalar triple product is zero.

### Step-by-Step Solution:

1. **Identify the Vectors:**
Let:
\[ \mathbf{a} = \hat{i} + 3\hat{j} + \hat{k} \quad (1) \]
\[ \mathbf{b} = 2\hat{i} - \hat{j} - \hat{k} \quad (2) \]
\[ \mathbf{c} = 0\hat{i} + 7\hat{j} + 3\hat{k} \quad (3) \]

2. **Set Up the Scalar Triple Product:**
The scalar triple product of vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) can be calculated using the determinant of a matrix formed by these vectors:
\[ \text{Scalar Triple Product} = \begin{vmatrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & 7 & 3 \end{vmatrix} \]

3. **Calculate the Determinant:**
We will expand the determinant along the first row:
\[ = 1 \cdot \begin{vmatrix} -1 & -1 \\ 7 & 3 \end{vmatrix} - 3 \cdot \begin{vmatrix} 2 & -1 \\ 0 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ 0 & 7 \end{vmatrix} \]

- Calculate the first 2x2 determinant:
\[ \begin{vmatrix} -1 & -1 \\ 7 & 3 \end{vmatrix} = (-1)(3) - (-1)(7) = -3 + 7 = 4 \]

- Calculate the second 2x2 determinant:
\[ \begin{vmatrix} 2 & -1 \\ 0 & 3 \end{vmatrix} = (2)(3) - (-1)(0) = 6 \]

- Calculate the third 2x2 determinant:
\[ \begin{vmatrix} 2 & -1 \\ 0 & 7 \end{vmatrix} = (2)(7) - (-1)(0) = 14 \]

4. **Substitute Back into the Determinant:**
Now substituting back into the determinant:
\[ = 1 \cdot 4 - 3 \cdot 6 + 1 \cdot 14 \]
\[ = 4 - 18 + 14 \]
\[ = 4 + 14 - 18 = 0 \]

5. **Conclusion:**
Since the scalar triple product is zero, we conclude that the vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are coplanar. Therefore, they are parallel to the same plane.