Final answer:
The distance from the point (2, 3, -2) to the plane 4x-4y-7z=-71 is calculated using the formula for the distance from a point to a plane and is found to be 9 units.
Step-by-step explanation:
The student is asking how to find the distance from a point to a plane. The point given is (2, 3, -2) and the equation of the plane is 4x-4y-7z=-71. The distance D from a point (x0, y0, z0) to a plane Ax+By+Cz+D=0 can be calculated using the formula:
D = |A*x0 + B*y0 + C*z0 + D| / √(A2 + B2 + C2)
Substituting the point into the equation, we get:
D = |4*2 - 4*3 - 7*(-2) - (-71)| / √(42 + (-4)2 + (-7)2)
D = |8 - 12 + 14 + 71| / √(16 + 16 + 49)
D = |81| / √(81)
D = 81 / 9
D = 9
Therefore, the distance from the point (2, 3, -2) to the plane 4x-4y-7z=-71 is 9 units.