Find the value of $\displaystyle a$ such that the planes $\displaystyle ax+y+z=0$, $\displaystyle x+3z=0$, and $\displaystyle 5y+6z=0$ have a line in common.
Let
$\displaystyle p_1:x+3z=0$
$\displaystyle p_2:5y+6z=0$ and
$\displaystyle p_3:ax+y+z=0$
denote the three planes.
1. Calculate
$\displaystyle p_1 \cap p_2 = l_{p_1,p_2} : \left\{\begin{array}{l}x=-3t \\y=-\frac65t \\z = t\end{array}\right.$
2. Calculate
$\displaystyle p_2 \cap p_3 = l_{p_2,p_3} : \left\{\begin{array}{l}x=\frac1{5a} t \\y=-\frac65t \\z = t\end{array}\right.$
3. Compare the coefficients:
$\displaystyle l_{p_1,p_2} = l_{p_2,p_3}\ if\ -3=\frac1{5a}~\implies~ a=\dfrac1{15}$
4. Thus the third plane has the equation:
$\displaystyle p_3:\dfrac1{15} x+y+z=0$
5. I've attached a sketch of the three planes.