Question

Consider three planes:
$2x+py+6z=8x+2y+qz=5x+y+3z=4$
The three planes do not have any common point of intersection if

• A. $p=2,q\ne 3$
• B. $p\ne 2,q\ne 3$
  
• C. $p\ne 2,q=3$
  
• D. $p=2,q=3$

Answer 1

The correct option is C $p\ne 2,q=3$


$A=\left[\begin{array}{ccc}2& p& 6\\ 1& 2& q\\ 1& 1& 3\end{array}\right]$

For the planes to not have any common point of intersection.
$A=0$ and at least any one $A_x,A_y,A_z$ is non-zero.

$A_z=\left[\begin{array}{ccc}2& p& 8\\ 1& 2& 5\\ 1& 1& 4\end{array}\right]$

$R_1\to R_1-2R_2$ and $R_2\to R_2-R_3$
$Az=\left[\begin{array}{ccc}0& p-4& -2\\ 0& 1& 1\\ 1& 1& 4\end{array}\right]$
$|A_z|\ne 0$

$\Rightarrow p-4+2\ne 0$
$\Rightarrow p\ne 2$

Also, $|A|=0$

$R_1\to R_1-2R_2$ and $R_2\to R_2-R_3$
$A=\left[\begin{array}{ccc}0& p-4& 6-2q\\ 0& 1& q-3\\ 1& 1& 3\end{array}\right]$
$\Rightarrow (p-4)(q-3)-(6-2q)=0$
$\Rightarrow pq-3p-2q+6=0$
$\Rightarrow p(q-3)=2(q-3)$
$\because p\ne 2\therefore q-3=0$
Hence, $q=3$