A) \[x+2y+4z=12\]
B) \[4x+2y+z=12\]
C) \[x+2y+4z=3\]
D) \[4x+2y+z=3\]
E) \[x+y+z=12\]
Correct Answer: B
Solution :
Let equation of plane is\[\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1,\]then A \[(\alpha ,0,0),B(0,\beta ,0)\]and\[C(0,0,\gamma )\]are the points on co-ordinate axes. Now, \[\frac{\alpha }{3}=1\] \[\Rightarrow \]\[\alpha =3,\frac{\beta }{3}=2\Rightarrow \beta =6\]and \[\frac{\gamma }{3}=4\Rightarrow \gamma =12\] \[\therefore \]Equation of plane is \[\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1\] \[\Rightarrow \] \[4x+2y+z=12\]
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