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...

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