I was hoping if someone could confirm if those 2 exercises are solved correctly.

1. we have a plane: (5x-y+z-2=0) and vector line in form: ((t)=x/-2=(y+1)/-3=(z-2)/1). What is the intersection between the two of them?

What i got is:

x=-2t

y=-3t-1

z=t+2

5(-2t)-(-3t-1)+t+2-2=0 t=1/6

x=-1/3

y=-3/2

z=13/6

2.find the perpendicular projection T' of point T(2,-1,5) on vector line : (x-2)/3=(1-z)/2 , y=-3. Also find a point what is a reflection of point T based on the projection T'.

What i did is:

s (3,0,-2) = n ( normal of plane that contains point T)

x=3t + 2

y=-3

z=1-2t

plane: 3x + 0y -2z = d ---> 3*2 -2*5=-4=d

3x-2z+4=0

3(3t+2)- 2(1-2t) +4 =0 ---> t = -8/13

T' (intersection between plane and vector line):

x= 3t+2=2/13

y=-3

z=1-2t=29/13

to get the point T'' i just calculated the difference between x,y,z coordinates of point T and T' and did the same for coordinates of T' and T'' so i got :

T''

x= 2/13 - 24/13 = -22/13

y= -3 -2 = -5

z = 29/13 - 36/13 = -7/13

so i got T'' (-24/13,-5,,-7/13)

Could this be correct? thank you