# Give an example of any plane that passes through point (-2,3,5) and is parallel to the line (3+5t, 2t, -2+3t). Express your answer in scalar...equation form.

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We'll have to determinte the normal vector to both planes. Since the vector n is perpendicular to the plane that contains the line (3+5t, 2t, -2+3t), then "n" is perpendicular to the line, too.

The coordinates of the given line are:

x = 3 + 5t

y = 2t

z = -2 + 3t

The vector of this line is: v = 5i + 2j + 3k

The n vector is: n = ai + bj + ck

The "n" vector is orthogonal to any vector that is located in the plane, therefore "n" is perpendicular to "v". Therefore, the dot product of n and v must be zero:

n*v = 0 <=> (ai + bj + ck)(5i + 2j + 3k) = 5a + 2b + 3c = 0 (1)

Since n is perpendicular to the plane that passes through the point (-2,3,5), then it is perpendicular to any vector that passe through this point, located in this plane.

Let's consider the vector p = -2i + 3j + 5k

Therefore, the dot product of n and p must be zero:

n*p = 0

(ai + bj + ck)(-2i + 3j + 5k) = -2a + 3b + 5c = 0 (2)

We'll find out the variable "c" from (1) and (2):

5a + 2b + 3c = 0 => c = (-5a+2b)/3 (3)

-2a + 3b + 5c = 0 => c = (2a-3b)/5 (4)

We'll equate (3) = (4):

(-5a+2b)/3 = (2a-3b)/5

We'll cross multiply:

6a - 9b = -25a + 10b

31a = 19b

a = 19b/31

c = (38b/31 - 3b)/5

c = (38b - 93b)/31*5

c = -55b/31*5

c = -11b/31

The coordinates of the vector n are: (19b/31 , b , -11b/31)

Let b = 31

n:(19 , 31 , -11)

The vector n is: n = 19i + 31j - 11k

**We can write the equation of the plane that passe through the point (-2,3,5) and it's parallel to the line(3+5t, 2t, -2+3t):**

**19(x + 2) + 31(y - 3) - 11(z - 5) = 0**