A) \[\frac{x}{0}=\frac{y}{0}=\frac{z+4}{1}\]
B) \[\frac{x}{1}=\frac{y}{2}=\frac{z}{-3}\]
C) \[\frac{x}{1}=\frac{y}{2}=\frac{z-4}{-7}\]
D) None of these
Correct Answer: C
Solution :
[c] Given plane is \[(2\lambda +1)x+(3-\lambda )y+z=4,(\lambda \in R)\] Or \[\lambda (2x-y)+x+3y+z-4=0\] This is family of planes containing the line of intersection of planes \[{{P}_{1}}:x+3y+z-4=0\] and \[{{P}_{2}}:2x-y=0\]. Let \[x=\alpha .\] \[\therefore \,\,\,\,\,\,y=2\alpha \] and \[z=4-7\alpha \] Thus, one of the lines on the plane is \[\frac{x}{1}=\frac{y}{2}=\frac{z-4}{-7}.\]
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