Answer:
[tex]x + 5y - 2z + 14 = 0[/tex]
Step-by-step explanation:
A plane that passes through the point [tex](x_{0}, y_{0}, z_{0})[/tex] with a normal vector of [tex](a,b,c)[/tex] has the following equation, initially:
[tex]a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0[/tex]
After we solve this, we have
[tex]ax + by + cz + d = 0[/tex]
In this problem, we have that:
[tex](x_{0}, y_{0}, z_{0}) = (-5,-1,2)[/tex]
[tex](a,b,c) = (−1,−5,2)[/tex]
So
[tex]a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0[/tex]
[tex]-1(x - (-5)) - 5(y - (-1)) + 2(z - 2) = 0[/tex]
[tex]-(x + 5) - 5(y + 1) + 2(z - 2) = 0[/tex]
[tex]-x - 5 - 5y - 5 + 2z - 4 = 0[/tex]
[tex]-x - 5y + 2z - 14 = 0[/tex]
Multplying everything by -1
[tex]x + 5y - 2z + 14 = 0[/tex]
