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# 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.

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