Find the shortest distance, d, from the point (2, 0, -5) to the plane x + y + z = 5. Step 1 The distance from (2, 0, -5) to a point (x, y, z) is given by d = (x Step 2 Step 3 For f(x, y) = (x - 2)² + y² + (10-x- y)², we get fx(x, y) = We will minimize d²2 = (x - 2)² + y² + (z + 5)². Since z = 5 - x - y, we can simplify this to d² = (x - 2)² + y² +1 fy(x, y) = )² + (y-0✔ and 0 )2 + (z+5✔ ². X 10-1-y

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