Passage:

Consider three planes 2x+py+6z=8, x + 2y + qz = 5 and x + y + 3z = 4

Question

Three planes do not have any common point of intersection if

Solution:

The given system of equations is 

$\begin{array}{l}2\mathrm{x}+\mathrm{py}+6\mathrm{z}=8\\ \mathrm{x}+2\mathrm{y}+\mathrm{qz}=5\\ \mathrm{x}+\mathrm{y}+3\mathrm{z}=4\\ \mathrm{\Delta }=\left|\begin{array}{lll}2& \mathrm{p}& 6\\ 1& 2& \mathrm{q}\\ 1& 1& 3\end{array}\right|=(2-\mathrm{p})(3-\mathrm{q})\end{array}$

$\text{ By Cramer's rule, if }\mathrm{\Delta }\ne 0\text{, i.e., }p\ne 2\text{ and }q\ne 3,\text{ the }$ system has a unique solution. 

$\text{ If }p=2\text{ or }q=3,\mathrm{\Delta }=0,\text{ then if }{\mathrm{\Delta }}_x={\mathrm{\Delta }}_v={\mathrm{\Delta }}_z=0,\text{ the }$system has infinite solutions and if any one of 

${\mathrm{\Delta }}_x,{\mathrm{\Delta }}_y\text{ and }{\mathrm{\Delta }}_z\ne 0$  the system has no solution

now,

$\begin{array}{l}{\mathrm{\Delta }}_{\mathrm{x}}=\left|\begin{array}{lll}8& \mathrm{p}& 6\\ 5& 2& \mathrm{q}\\ 4& 1& 3\end{array}\right|\\ =30-8\mathrm{q}-15\mathrm{p}+4\mathrm{pq}=(4\mathrm{q}-15)\cdot (\mathrm{p}-2)\\ =\left|\begin{array}{lll}2& 8& 6\\ 1& 5& \mathrm{q}\\ 1& 4& 3\end{array}\right|\\ =-8\mathrm{q}+8\mathrm{q}=0\\ {\mathrm{\Delta }}_{\mathrm{y}}=\left|\begin{array}{lll}2& \mathrm{p}& 8\\ 1& 2& 5\\ 1& 1& 4\end{array}\right|\\ =\mathrm{p}-2\end{array}$

Thus, if $\mathrm{p}=2,{\mathrm{\Delta }}_{\mathrm{x}}={\mathrm{\Delta }}_{\mathrm{y}}={\mathrm{\Delta }}_{\mathrm{z}}=0$  for all $\mathrm{q}\in \mathrm{R}$  the system has infinite solutions. 

if $\mathrm{p}\ne 2,\mathrm{q}=3\text{ and }{\mathrm{\Delta }}_{\mathrm{z}}\ne 0$  then the system has no solution. 

Hence the system has (i) no solution if $\mathrm{p}\ne 2$  and q = 3,(ii) a unique solution if $\mathrm{p}\ne 2$ and $\mathrm{q}\ne 2$ and (iii) infinite solutions if p =  2 and $\mathrm{q}\in \mathrm{R}$