Find the equation of the line in parametric form that is parallel to the line \( [x, y, z]=[4-3 t,-2-2 t,-5-4 t] \) and has an intercept at \( x=3 \). \[ \begin{array}{ll}x=\text { Ex: } 6+8 \mathrm{t} \\ y= \\ z= & \text { Enter as: } a+b t \text { or } a-b t\end{array} \]

To find the equation of the line in parametric form that is parallel to the given line \( [x, y, z] = [4 - 3t, -2 - 2t, -5 - 4t] \) and has an intercept at \( x = 3 \), we first identify the direction vector of the given line, which is determined from the coefficients of \( t \). The direction vector is \([-3, -2, -4]\). Now, since the new line is parallel, it will have the same direction vector. However, we need to adjust it so that \( x = 3 \) when \( t = 0 \). This leads us to set up the equations as follows: 1. For the new line, at \( t = 0 \): \[ x = 3 \] Thus, we can express \( x \) in parametric form: \[ x = 3 - 3t \] (We chose \( -3t \) to keep the same direction as the original.) 2. Since it is a parallel line, we can write: \[ y = y_0 - 2t \] where \( y_0 \) is the \( y \)-intercept. For simplicity, we can choose \( y_0 = 0 \): \[ y = 0 - 2t = -2t \] 3. For the \( z \) value, we similarly set: \[ z = z_0 - 4t \] where \( z_0 \) can also be chosen conveniently; let’s take \( z_0 = 0 \): \[ z = 0 - 4t = -4t \] Now we have the parametric equations: \[ \begin{array}{ll} x = 3 - 3t \\ y = -2t \\ z = -4t \end{array} \] Rearranging the equations gives us the final parametric form of the line: \[ \begin{array}{ll} x = 3 - 3t \\ y = -2t \\ z = -4t \end{array} \] Therefore, the results can be entered as: \[ x=3-3t, \quad y=-2t, \quad z=-4t \]