The problem asks us to find the value of $y$ such that the three vectors $\vec{a} = 7\hat{i} + y\hat{j} + \hat{k}$, $\vec{b} = 3\hat{i} + 2\hat{j} + \hat{k}$, and $\vec{c} = 5\hat{i} + 3\hat{j} + \hat{k}$ are linearly dependent.

1. Problem Description

The problem asks us to find the value of $y$ such that the three vectors $\vec{a} = 7\hat{i} + y\hat{j} + \hat{k}$, $\vec{b} = 3\hat{i} + 2\hat{j} + \hat{k}$, and $\vec{c} = 5\hat{i} + 3\hat{j} + \hat{k}$ are linearly dependent.

2. Solution Steps

Three vectors are linearly dependent if their scalar triple product is zero. The scalar triple product of $\vec{a}$, $\vec{b}$, and $\vec{c}$ is given by the determinant of the matrix formed by their components:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 7 & y & 1 \\ 3 & 2 & 1 \\ 5 & 3 & 1 \end{vmatrix}$

For the vectors to be linearly dependent, we require

$\begin{vmatrix} 7 & y & 1 \\ 3 & 2 & 1 \\ 5 & 3 & 1 \end{vmatrix} = 0$

Expanding the determinant, we get:

$7(2 \cdot 1 - 1 \cdot 3) - y(3 \cdot 1 - 1 \cdot 5) + 1(3 \cdot 3 - 2 \cdot 5) = 0$

$7(2 - 3) - y(3 - 5) + (9 - 10) = 0$

$7(-1) - y(-2) + (-1) = 0$

$-7 + 2y - 1 = 0$

$2y - 8 = 0$

$2y = 8$

$y = 4$

3. Final Answer

The value of $y$ that makes the vectors linearly dependent is $y = 4$.