Find the point (if it exists) at which the following plane and line intersect. \( -2 x+6 y+2 z=-15 \) and \( x=-9-4 t, y=-3+4 t, z=4-10 t \) Select the correct choice below and, if necessary, fill in any answer boxes to complete your choice. A. The point at which the line intersects the plane is (Simplify your answers. Type an ordered triple.) B. The line and the plane do not intersect.

To find the intersection of the line and the plane, we can substitute the parametric equations of the line into the equation of the plane. Given the plane's equation: \[ -2x + 6y + 2z = -15 \] And the parametric equations for the line: \[ x = -9 - 4t,\quad y = -3 + 4t,\quad z = 4 - 10t \] Substituting the values of \(x\), \(y\), and \(z\) from the line into the plane's equation: \[ -2(-9 - 4t) + 6(-3 + 4t) + 2(4 - 10t) = -15 \] Calculating this step-by-step: 1. Solve for \( -2(-9 - 4t) = 18 + 8t \) 2. Solve for \( 6(-3 + 4t) = -18 + 24t \) 3. Solve for \( 2(4 - 10t) = 8 - 20t \) Now, combine these: \[ 18 + 8t - 18 + 24t + 8 - 20t = -15 \] This simplifies to: \[ 12t + 8 = -15 \] Solving for \(t\): \[ 12t = -23 \quad \Rightarrow \quad t = -\frac{23}{12} \] Now plug \(t\) back into the parametric equations to find the intersection point: \[ x = -9 - 4\left(-\frac{23}{12}\right) = -9 + \frac{92}{12} = -9 + \frac{23}{3} = -\frac{27}{3} + \frac{23}{3} = -\frac{4}{3} \] \[ y = -3 + 4\left(-\frac{23}{12}\right) = -3 - \frac{92}{12} = -\frac{36}{12} - \frac{92}{12} = -\frac{128}{12} = -\frac{32}{3} \] \[ z = 4 - 10\left(-\frac{23}{12}\right) = 4 + \frac{230}{12} = \frac{48}{12} + \frac{230}{12} = \frac{278}{12} = \frac{139}{6} \] Thus, the point at which the line intersects the plane is: \[ \left(-\frac{4}{3}, -\frac{32}{3}, \frac{139}{6}\right) \]