Question

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

Answer: (C)

$p\ne 2,q=3$

Given planes $2x+py+6z=8,x+2y+qz=5$ and $x+y+3z=4$ have no common point of intersection.
$\Rightarrow$ given system of equations has no solution.
Augumented matrix is $\left[\begin{array}{cccc}2& p& 6& 8\\ 1& 2& q& 5\\ 1& 1& 3& 4\end{array}\right]$
Interchanging $R_1$ and $R_3$
$\left[\begin{array}{cccc}1& 1& 3& 4\\ 1& 2& q& 5\\ 2& p& 6& 8\end{array}\right]$
Applying $R_3\to R_3-2R_1$ and $R_2\to R_2-R_1$
$\left[\begin{array}{cccc}1& 1& 3& 4\\ 0& 1& q-3& 1\\ 0& p-2& 0& 0\end{array}\right]$
If $p=2$ system has infinitely many solutions.
$\therefore p\ne 2$
$\Rightarrow$ from $R_3$, $y=0$
and if $q=3$ from $R_2$, $y=1$
$\therefore$ given system has no solution if $p\ne 2$ and $q=3$.
Hence, option C.