Finding the value of K

Question

$$π1:( 1+k , -3 , 6 )$$

$$π2: ( 1 , 5+k , 3k)$$

Does anyone know the values of k that would make these parallel and perpendicular, ive been trying for hours but nothing seems to work

The original equation is

$$π1:(1+k)x−3y+6z−4=0$$

$$π2:x+(5+k)y+3kz+1=0$$

Answer 1

Hints:

• If $v_1$ and $v_2$ are perpendicular, $v_1 \cdot v_2 = 0$.
• If $v_1$ and $v_2$ are parallel, $v_2 = av_2$ for some constant $a$.