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

# Passage:Consider three planes 2x+py+6z=8, x + 2y + qz = 5 and x + y + 3z = 4QuestionThree planes do not have any common point of intersection if

1. A

$\mathrm{p}=2,\mathrm{q}\ne 3$

2. B

$\mathrm{p}\ne 2,\mathrm{q}\ne 3$

3. C

$\mathrm{p}\ne 2,\mathrm{q}=3$

4. D

$\mathrm{p}=2,\mathrm{q}=3$

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### Solution:

The given system of equations is

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

system has a unique solution.

system has infinite solutions and if any one of

the system has no solution

now,

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

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   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 $\mathrm{q}\in \mathrm{R}$

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