Find an equation of the plane through the point (−1,−5,−2)(−1,−5,−2) with normal vector n=⟨−5,−4,−1⟩n=⟨−5,−4,−1⟩.

Consider this option:
1. given: n(A;B;C)=n(-5;-4;-1) and A(-1;-5;-2)=A(x₀;y₀;z₀).
Common view of equation for a plane is Ax+By+Cz+D=0, where A,B,C,D - numbers.
2. from another side using coordinates of A and normal vector it is possible to make up the equation: A(x-x₀)+B(y-y₀)+C(z-z₀)=0 ⇒ -5(x+1)-4(y+5)-(z+2)=0; ⇒ -5x-4y-z-27=0 or 5x+4y+z+27=0.

128/3reason being 5 and 1/3 = 16/316/3 • 2/3 = 32/31 and 1/3 = 4/3 32/3 • 4/3 this will equal 128/3