The distance D from a point P(x₁, y₁, z₁) to a plane ax + by + cz = d can be given by the formula:
D = abs(ax₁ + by₁ + cz₁ - d) / (sqrt(a² + b² + c²))
We can use this formula with our given point and plane.
When we substitute Point P(-4, -1, 0) into the formula:
ax₁ + by₁ + cz₁ - d = 5(-4) - 2(-1) - 6(0) - 3
This simplifies to -20 + 2 - 3, that provides -21
And a² + b² + c² = 5² + (-2)² + (-6)² = 25 + 4 + 36 = 65
So,
D = |ax₁ + by₁ + cz₁ - d| / sqrt(a² + b² + c²) = |-21| / sqrt(65) = 21 / sqrt(65)
Using a square root calculator, sqrt(65) equals approximately 8.062. If we take 21 / 8.062, it equals approximately 2.604.
So, the distance from the point P(-4, -1, 0) to the plane 5x - 2y - 6z = 3 is approximately 2.604 units. But among given four closest options, it is closer to option b) 4 units.