Distance of a point to a Plane

Calculate distance of a point (3,7,-4) to the Plane 6x-3y+2z = 10

Let the equation of the plane :
w1.i + w2.j + w3.k + w0 = 0

Then, W = w1.i + w2.j + w3.k be the perpendicular vector to the plane

vector(b) = vector(X)-vector(P)
Let X = a.i + b.j + c.k
and P = p1.i + p2.j+p3.k

then, vector(b) = (a-p1).i + (b-p2).j + (c-p3).k

Angle(PXO) = Angle(XPW) = Θ

And, CosΘ = d/b (since, base/hypotenuse= cosΘ)
d = b*CosΘ

Vector(W)*Vector(b) = W*b*CosΘ (Vector Dot Product)

w1(a-p1) + w2(b-p2) + w3(c-p3) = W*d

W*d = w1*a + w2*b + w3*c -(w1*p1 + w2*p2 + w3*p3)

W*d = w1*a + w2*b + w3*c -(-w0)

d = (w1*a + w2*b + w3*c + w0)/W
d = (W *(dot_Product)X + w0) / Sqrt(w1² + w2² + w3²)

Lets solve this ‘Calculate distance of a point (3,7,-4) to the Plane 6x-3y+2z = 10’

Distance, d =( 6*3+3*(-7)+2*(-4)+(–10) ) / Sqrt(6²+3² + 2²)

d = -21/7 = -3