Find the equation for the plane through the points Upper P 0 (5 comma 4 comma 5 ), Upper Q 0 (negative 5 comma negative 1 comma negative 4 ), and Upper R 0 (negative 2 comma 1 comma negative 2 ). The equation of the plane is nothing.

The equation of the plane is

8(x - 5) - 7(y - 4) - 5(z - 5) = 0

8x - 7y - 5z + 13 = 0

Step-by-step explanation:

Given 3 points, P(x₁, y₁, z₁), Q(x₂, y₂, z₂), and R(x₃, y₃, z₃).

We can calculate the equation of the plane through those points as

a(x - x₀) + b(y - y₀) + c(z - z₀) = 0, where (x₀, y₀, z₀) are the coordinates of any one of the points P, Q, or R, and <a,b,c> is a vector perpendicular to the plane.

The vector perpendicular to the plane is obtained by writing vector PQ and PR and taking the cross or vector product.

For this question,

P = (5, 4, 5)

Q = (-5, -1, -4)

R = (-2, 1, -2)

PQ = (-5, -1, -4) - (5, 4, 5) = (-10, -5, -9)

= (-10î - 5ĵ - 9ķ)

PR = (-2, 1, -2) - (5, 4, 5) = (-7, -3, -7)

= (-7î - 3ĵ - 7ķ)

PQ × PR is then

| î ĵ ķ |

|-10 -5 -9|

|-7 -3 -7|

= î [(-5×-7) - (-9×-3)] - ĵ [(-10×-7) - (-9×-7)] + ķ [(-10×-3) - (-7×-5)]

= î (35 - 27) - ĵ (70 - 63) + ķ (30 - 35)

= 8î - 7ĵ - 5ķ

Hence, (a, b, c) = (8, -7, -5)

And using point P as (x₀, y₀, z₀) = (5, 4, 5)

The equation of the plane is

a(x - x₀) + b(y - y₀) + c(z - z₀) = 0

8(x - 5) - 7(y - 4) - 5(z - 5) = 0

8x - 40 - 7y + 28 - 5z + 25 = 0

8x - 7y - 5z = 40 - 28 - 25 = -13

8x - 7y - 5z + 13 = 0

Hope this Helps!!!