### Video Transcript

Find the distance between the point
two, eight, three and the plane 𝑥 minus two plus two times 𝑦 minus one plus five
times 𝑧 minus four equals zero to the nearest two decimal places.

In this question, we want to
determine the distance between a given point and plane to the nearest two decimal
places. To do this, we can start by
recalling that when we say the distance between a point and a plane, we mean the
perpendicular distance since it is the shortest distance between the two
objects. We can find the distance between
the point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one and the plane 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧
plus 𝑑 equals zero using the formula the absolute value of 𝑎𝑥 sub one plus 𝑏𝑦
sub one plus 𝑐𝑧 sub one plus 𝑑 all divided by the square root of 𝑎 squared plus
𝑏 squared plus 𝑐 squared.

We are given the coordinates of the
point in the question. These coordinates tell us the
values of 𝑥 sub one, 𝑦 sub one, and 𝑧 sub one. However, we are not given the
equation of the plane in the correct form. We can rewrite this equation in the
correct form by expanding and simplifying. First, we distribute each factor
over the parentheses to obtain 𝑥 minus two plus two 𝑦 minus two plus five 𝑧 minus
20 is equal to zero. Second, we evaluate the constant
terms in the equation to get 𝑥 plus two 𝑦 plus five 𝑧 minus 24 equals zero.

Now, the equation of the plane is
in the correct form. So the coefficients of 𝑥, 𝑦, and
𝑧 will tell us the values of 𝑎, 𝑏, and 𝑐, and the constant tells us the value of
𝑑. We have 𝑎 equals one, 𝑏 equals
two, 𝑐 equals five, and 𝑑 equals negative 24.

We can now find the distance
between the point and the plane by substituting these values into the formula and
evaluating. We get that the distance is the
absolute value of one times two plus two times eight plus five times three minus 24
all over the square root of one squared plus two squared plus five squared. Evaluating this expression gives us
nine divided by the square root of 30. We can evaluate this radical using
a calculator to obtain 1.643, and this expansion continues. Rounding this to two decimal places
gives us that the distance between the point and the plane is 1.64 to the nearest
two decimal places.