A) \[x-y+z=1\]
B) \[x+y+z=5\]
C) \[x+2y-z=1\]
D) \[2x-y+z=5\]
Correct Answer: A
Solution :
[a] The required plane is \[\left| \begin{matrix} x-3 & y-6 & z-4 \\ 3-3 & 2-6 & 0-4 \\ 1 & 5 & 4 \\ \end{matrix} \right|=0\] Or \[\left| \begin{matrix} x-3 & y-z-2 & z-4 \\ 0 & 0 & -\,4 \\ 1 & 1 & 4 \\ \end{matrix} \right|=0\] (Operating\[{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\]) Or \[4(x-3-y+z+2)=0\] Or \[x-y+z=1\]
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