Step 1: Understanding the Question:
The question asks for the value of 'a' that makes three given vectors lie on the same plane. Vectors that lie on the same plane are called coplanar vectors.
Step 2: Key Formula or Approach:
Three vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) are coplanar if and only if their scalar triple product is zero. The scalar triple product can be calculated as the determinant of the matrix formed by their components.
\[
[\vec{A} \vec{B} \vec{C}] = \vec{A} \cdot (\vec{B} \times \vec{C}) = \begin{vmatrix} A_x & A_y & A_z
B_x & B_y & B_z
C_x & C_y & C_z \end{vmatrix} = 0
\]
Step 3: Detailed Explanation:
Let the given vectors be:
\(\vec{A} = 2\hat{i} - \hat{j} + \hat{k}\)
\(\vec{B} = \hat{i} + 2\hat{j} - 3\hat{k}\)
\(\vec{C} = 3\hat{i} + a\hat{j} + 5\hat{k}\)
For the vectors to be coplanar, the determinant of their components must be zero:
\[
\begin{vmatrix}
2 & -1 & 1
1 & 2 & -3
3 & a & 5
\end{vmatrix} = 0
\]
Now, we expand the determinant along the first row:
\[
2 \begin{vmatrix} 2 & -3
a & 5 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -3
3 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2
3 & a \end{vmatrix} = 0
\]
\[
2((2)(5) - (-3)(a)) + 1((1)(5) - (-3)(3)) + 1((1)(a) - (2)(3)) = 0
\]
\[
2(10 + 3a) + 1(5 + 9) + 1(a - 6) = 0
\]
\[
20 + 6a + 14 + a - 6 = 0
\]
Combine the terms:
\[
7a + 28 = 0
\]
Solve for \(a\):
\[
7a = -28
\]
\[
a = -4
\]
Step 4: Final Answer:
The value of \(a\) for which the vectors are coplanar is \(-4\).