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 \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) linearly dependent, we can set up a matrix using the components of the vectors and compute the determinant. If the determinant of this matrix is zero, the vectors are linearly dependent. The vectors can be represented in a matrix as follows: \[ \begin{bmatrix} 7 & 3 & 5 \\ y & 2 & 3 \\ 1 & 1 & 1 \end{bmatrix} \] We need to compute the determinant of this matrix and set it to zero: \[ \text{det} = 7 \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} - 3 \begin{vmatrix} y & 3 \\ 1 & 1 \end{vmatrix} + 5 \begin{vmatrix} y & 2 \\ 1 & 1 \end{vmatrix} \] Calculating the 2x2 determinants: \[ \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = (2)(1) - (3)(1) = 2 - 3 = -1 \] \[ \begin{vmatrix} y & 3 \\ 1 & 1 \end{vmatrix} = (y)(1) - (3)(1) = y - 3 \] \[ \begin{vmatrix} y & 2 \\ 1 & 1 \end{vmatrix} = (y)(1) - (2)(1) = y - 2 \] Now substituting these back into the determinant equation: \[ \text{det} = 7(-1) - 3(y - 3) + 5(y - 2) \] Expanding this: \[ \text{det} = -7 - 3y + 9 + 5y - 10 \] \[ = (5y - 3y) + (-7 + 9 - 10) \] \[ = 2y - 8 \] Setting the determinant equal to zero for linear dependence: \[ 2y - 8 = 0 \] Solving for \( y \): \[ 2y = 8 \\ y = 4 \] Thus, the value of \( y \) that makes the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) linearly dependent is \( y = 4 \).