Find the shortest distance d, from the point (3, 0, -2) to the plane x + y + z = 2.

Solution:

Given the point (3, 0, -2)

The plane x + y + z = 2 ⇒ (1, 1, 1)

The shortest distance of a plane from the point can be calculated by

√{(x\(_2\) - x\(_1\))² + (y\(_2\) - y\(_1\))² + (z\(_2\) - z\(_1\))²}

= √{(3 - 1)² + (0 - 1)² + (-2 - 1)²}

= √{(2)² + (-1)² + (-3)²}

= √{4 + 1 + 9}

= √14

Find the shortest distance d, from the point (3, 0, -2) to the plane x + y + z = 2.

Summary:

The shortest distance d, from the point (3, 0, -2) to the plane x + y + z = 2 is √14 units.