Find an equation of a plane through the point (3, 0, 2) which is orthogonal to the linex=?2+3t,y=2+4t,z=?1?3tin which the coefficient of x is 3.

3x + 4y - 3z - 3 = 0

Step-by-step explanation:

with the given point (3, 0, 2), the plane is orthogonal to this line so it

has directional ratios (3, 4, -3)

therefore the given equation can be written as;

(x+2)/3 = (y-2)/4 = (z+1)/-3

so equation of the plane passing through point x1, y1 and z1 in the one point form is given as;

a(x-x1) + b(y - y1) + c(z -z1) = 0

abc represent the direction ratio

so we substitute

3(x-3) + 4(y-0) + (-3(z-2)) = 0

3x - 9 + 4y - 0 - 3z + 6 = 0

3x + 4y - 3z - 3 = 0

therefore the equation of the plane through the point is  3x + 4y - 3z - 3 = 0

answer: 6(r+2)(r−10)

Step-by-step explanation: