Find the intersection of the plane \( 3 x-2 y-7 z=-30 \) and the line \( \vec{r}(t)=\langle-4,-2,4\rangle t+\langle 5,-7,-7\rangle \). \( y= \) \( z= \) Submit Question

To find the intersection of the plane \( 3x - 2y - 7z = -30 \) and the line defined by \( \vec{r}(t) = \langle -4, -2, 4 \rangle t + \langle 5, -7, -7 \rangle \), we can start by expressing the line in parametric form: \[ x = -4t + 5, \] \[ y = -2t - 7, \] \[ z = 4t - 7. \] Next, we substitute these expressions into the plane equation: \[ 3(-4t + 5) - 2(-2t - 7) - 7(4t - 7) = -30. \] Now, simplifying: \[ -12t + 15 + 4t + 14 - 28t + 49 = -30, \] \[ -36t + 78 = -30. \] Solving for \( t \): \[ -36t = -30 - 78, \] \[ -36t = -108, \] \[ t = 3. \] Now, substitute \( t = 3 \) back into the parametric equations to find \( x, y, \) and \( z \): \[ x = -4(3) + 5 = -12 + 5 = -7, \] \[ y = -2(3) - 7 = -6 - 7 = -13, \] \[ z = 4(3) - 7 = 12 - 7 = 5. \] Thus, the intersection point is: \[ y = -13, \] \[ z = 5. \]