Normal equation of view plane?

Question

Answer 1

(1).(a).

The plane equation is ax + by + cz + d = 0, where <a, b, c> is normal vector to the plane.

The normal vector n = (a, b, c) = (1, -3, 4).

The plane equation is x - 3y + 4z + d = 0.

Find the constant value d by substituing (1, -3, 4) in the equation.

(1)(1)-(3(-3)+4(4)+d=0

1+9+16+d=0

d=-26

 x - 3y + 4z - 26 = 0.

The plane equation is x - 3y + 4z - 26 = 0

Comments

Does 'd' always have to be in the equation?

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If given another point passing through the plane then only we can eliminate d or if  any another condition is given then only we can eliminate d.

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we were told u.n = n.n therefore d = n.n and our equations are in form ax+by+cz=d so x-3y+4z=26

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Solution is updated please check

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Thank you :)

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How do i work out the second part of this?