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The problem asks for the distance from the point \((1, 0, -1)\) to the plane given by the equation:

\[
2x - y + z = 7
\]

The formula to find the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane of the form \(Ax + By + Cz + D = 0\) is:

\[
d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]

First, we rewrite the plane equation in the form \(2x - y + z - 7 = 0\), so \(A = 2\), \(B = -1\), \(C = 1\), and \(D = -7\).

Given the point \((1, 0, -1)\), we substitute into the formula:

\[
d = \frac{|2(1) - 1(0) + 1(-1) - 7|}{\sqrt{2^2 + (-1)^2 + 1^2}}
\]

Now let's compute the exact distance.The distance from the point \((1, 0, -1)\) to the plane \(2x - y + z = 7\) is approximately \(2.45\) units.

Would you like more details on how the formula works or have any further questions?

Here are some related questions:
1. How do you derive the distance formula for a point to a plane?
2. What are the conditions for a point to lie on the plane?
3. Can you find the perpendicular projection of the point onto the plane?
4. How does the distance between a point and a plane change with different plane orientations?
5. What if the plane equation was given in a different form, such as parametric form?

**Tip**: When working with vectors and planes, understanding how the normal vector relates to the plane's orientation can be quite helpful!

La distance du point (1, 0, -1) au plan d'équation 2x - y + z = 7 est:

Learn how to find the distance from the point (1, 0, -1) to the plane 2x - y + z = 7 using the distance formula. This problem is suited for students in Grades 10-12 and covers key geometry concepts.

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