# Parallels.Prove that the line d is parallel to the plane P if the equation of the plane is 2x+3y+z-1=0.

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How can anyone say that the given Plane P which is 2x+3y+z-1=0 is parallel to the given line d when the equation of the line is not provided. Unless the equation of the line is provided it is not possible to determine if it is parallel to the plane or not.

I do not know where the previous editor found the parametric equation of the line, that has been used in the response, from.

The parametric equations of the line are x=1+2t, y=2-3t, z=5t

The line d is parallel to the plane P, if and only if the vector parallel to the line, v, and the vector perpendicular to the plane P, n, are perpendicular.

Two vectors are perpendicular if and only if their dor product is zero:

v*n = 0

Since the equation of the plane P is given, we'll identify the parametric coefficients of the normal vector to the plane, n.

2x+3y+z-1=0, where n(2 , 3 , 1)

We'll write n = 2i + 3j + k

Since the position vector of the line d is:

r = r0 + t*v

We'll write the parametric equations of d:

x = x0 + t*vx

y = y0 + t*vy

z = z0 + t*vz

Comparing the given parametric equations and the general parametric equations, we'll identify the parametric coefficients of the vector v:

x=1+2t

y=2-3t

z=5t

v (2 , -3 , 5)

v = 2i - 3j + 5k

Now, we'll write the dot product of n and v:

n*v = 2*2 + 3*(-3) + 1*5

If n*v = 0, the line d is parallel to the plane P.

n*v = 4 - 9 + 5

n*v = 0

**Since n*v = 0, the given line is parallel to the plane P.**