Verified answer

Distance from point (x₀, y₀, z₀) to plane Ax + By + Cz + D = 0

d = |Ax₀ + By₀ + Cz₀ + D| / √(A² + B² + C²)

Distance from point (8, 0, −6) to plane x + y + z − 6 = 0

d = |8 + 0 − 6 − 6| / √(1² + 1² + 1²)

d = 4/√3

For (8,0,-6), we have d^2=distance square is (x-8)^2+(y-0)^2+(z+6)^2. Use the Lagrange multiplier L. You have'

grad ( (x-8)^2+(y-0)^2+(z+6)^2)= L grad(x+y+z-6). You have equation

2(x-8)=L, 2y=L, 2(z+6)=L. Add akk of them. You get

2(x+y+z) -16+12=3L, 2*6-16+12=3L=8, L=8/3,

This gives y=4/3

x=17/6, and z=11/6. The shortest distance =sqrt((x-8)^2+y^2+(z+6)^2). When we put blues of x, y and z, we that is 5.65. Please check the answer.