3. Choose \( y \) in order that the vectors \( \vec{a}=7 \hat{i}+y \hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}+\hat{k} \), \( \vec{c}=5 \hat{i}+3 \hat{j}+\hat{k} \) are linearly dependent.

To determine the value of \( y \) that makes the vectors \( \vec{a} = 7\hat{i} + y\hat{j} + \hat{k} \), \( \vec{b} = 3\hat{i} + 2\hat{j} + \hat{k} \), and \( \vec{c} = 5\hat{i} + 3\hat{j} + \hat{k} \) linearly dependent, we can use the concept of the determinant of a matrix formed by these vectors. ### Step 1: Form the Matrix Arrange the vectors as columns in a matrix: \[ \begin{pmatrix} 7 & 3 & 5 \\ y & 2 & 3 \\ 1 & 1 & 1 \end{pmatrix} \] ### Step 2: Calculate the Determinant For the vectors to be linearly dependent, the determinant of this matrix must be zero. \[ \text{Determinant} = 7 \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} - 3 \begin{vmatrix} y & 3 \\ 1 & 1 \end{vmatrix} + 5 \begin{vmatrix} y & 2 \\ 1 & 1 \end{vmatrix} \] Calculate each 2x2 determinant: \[ \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = (2)(1) - (3)(1) = -1 \] \[ \begin{vmatrix} y & 3 \\ 1 & 1 \end{vmatrix} = y(1) - 3(1) = y - 3 \] \[ \begin{vmatrix} y & 2 \\ 1 & 1 \end{vmatrix} = y(1) - 2(1) = y - 2 \] Substitute back into the determinant: \[ \text{Determinant} = 7(-1) - 3(y - 3) + 5(y - 2) = -7 - 3y + 9 + 5y - 10 \] Combine like terms: \[ -7 + 9 - 10 + (-3y + 5y) = -8 + 2y \] ### Step 3: Set the Determinant to Zero \[ -8 + 2y = 0 \implies 2y = 8 \implies y = 4 \] **Answer:** \( y = 4 \)