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The distance formula is a valuable tool in geometry and algebra to determine the shortest path between a point and a geometric entity. In the context of finding the distance between a point and a plane, the formula is slightly more complex than the familiar coordinate distance calculation. It involves a substitution of the given point into the plane equation and the normalisation of the plane's coefficients.

The formula looks like: \[ d = \frac{ |ax_1 + by_1 + cz_1 + d| }{ \sqrt{a^2 + b^2 + c^2} } \]
Here, \((x_1, y_1, z_1)\) represents the coordinates of the point, and the plane's equation is expressed as \(ax + by + cz + d = 0\). The numerator involves the absolute value of the expression obtained by substituting the point into the plane's equation, ensuring a positive distance. The denominator is the square root of the sum of the squares of the plane's coefficients, which normalises the equation.

In three-dimensional space, a plane can be described mathematically by an equation of the form
\[ax + by + cz + d = 0\]
where \(a\), \(b\), and \(c\) are coefficients that determine the orientation of the plane in space by representing the components of the plane's normal vector, and \(d\) is the scalar distance from the origin to the plane along the normal vector. This equation encapsulates all the points \((x, y, z)\) that lie on the plane. When we seek the distance of a specific point to the plane, we use these coefficients in conjunction with the point's coordinates in the distance formula.

The substitution method plays a critical role when applying the distance formula. This method involves replacing the variables in the plane's equation with the respective coordinates of the given point to evaluate how 'far' the point deviates from satisfying the equation—essentially measuring its perpendicular distance from the plane.

In the given exercise, we substituted the point \((2, 8, 4)\) into the plane's equation \(2x + y + z = 5\) and computed \(2*2 + 8 + 4 - 5\), which is part of the distance formula's numerator.

The square root function is fundamental in the distance formula as it's involved in calculating the denominator. Specifically, it is used to normalize the coefficients of the plane's equation. In the provided exercise, we calculate the square root of the sum of the squares of the coefficients \(a\), \(b\), and \(c\) to find the magnitude of the plane's normal vector, which is \(\sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}\).

This step ensures that we account for the three-dimensional aspect of the plane when quantifying the distance from a point. Without the square root providing the proper scale, our distance measurement would be meaningless.