Find the point at which the line \( \langle 3,1,4\rangle+t\langle 2,-1,1\rangle \) intersects the plane \( -2 x+3 y+z=-11 \).

To find the intersection of the line given by \( \langle 3,1,4\rangle+t\langle 2,-1,1\rangle \) and the plane defined by the equation \( -2x + 3y + z = -11 \), we can substitute the line's parametric equations into the plane's equation. The parametric equations of the line can be expressed as follows: - \( x = 3 + 2t \) - \( y = 1 - t \) - \( z = 4 + t \) Now substituting these into the plane equation: \[ -2(3 + 2t) + 3(1 - t) + (4 + t) = -11 \] Expanding this out: \[ -6 - 4t + 3 - 3t + 4 + t = -11 \] Combining like terms: \[ (-4t - 3t + t) + (-6 + 3 + 4) = -11 \] This simplifies to: \[ -6t + 1 = -11 \] Now, solving for \( t \): \[ -6t = -12 \\ t = 2 \] Now that we have \( t \), we can find the coordinates of the intersection point by substituting \( t = 2 \) back into the parametric equations of the line: - \( x = 3 + 2(2) = 3 + 4 = 7 \) - \( y = 1 - 2 = -1 \) - \( z = 4 + 2 = 6 \) Thus, the point of intersection is: \[ \boxed{(7, -1, 6)} \]