Question

Consider three planes:
$2x+py+6z=8x+2y+qz=5x+y+3z=4$
The three planes intersect at a point 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 B $p\ne 2,q\ne 3$


Given planes are:

$2x+py+6z=8x+2y+qz=5x+y+3z=4$

Augmented matrix, $A=\left[\begin{array}{cccc}2& p& 6& 8\\ 1& 2& q& 5\\ 1& 1& 3& 4\end{array}\right]$

We will find the rank of matrix $A$. For the planes to intersect in a point, the rank of this matrix must be 3.

$R_3\to R_3-\frac{1}{2}{R}_1$ and $R_2\to R_2-\frac{1}{2}{R}_1$

$A=\left[\begin{array}{cccc}2& p& 6& 8\\ 0& 2-\frac{p}{2}& q-3& 1\\ 0& 1-\frac{p}{2}& 0& 0\end{array}\right]$

For the given system of equation to have unique solution,
$1-\frac{p}{2}\ne 0$ and $q-3\ne 0$
$\Rightarrow p\ne 2$ and $q\ne 3$