A) It lies in the plane \[x-2y\text{ }+\text{ }z=0\]
B) It is same as line \[\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\]
C) It passes through (2, 3, 5)
D) It is parallel to the plane \[x-2y+\text{ }z-6=0\]
Correct Answer: C
Solution :
On (1,2,3) satisfies the plane \[x-2y+z=0\] & also \[(\hat{i}+2\hat{j}+3\hat{k}).\]\[(\hat{i}-2\hat{j}+\hat{k})=0\] Since, the lines \[\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\] and \[\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\] both satisfy (0,0,0) & (1,2,3), both are same Given line is obviously parallel to the plane \[x-2y+z=6\]
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