A) It lines in the plane \[x-2y+z=0\]
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 of the plane \[x-2y+z-6=0\]
Correct Answer: C
Solution :
[c] (1, 2, 3) satisfies the plane \[x-2y+z=0\] and \[(\hat{i}+2\hat{j}+3\hat{k})\cdot (\hat{i}-2\hat{j}+\hat{k})=0\] 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) and (1, 2, 3) both are same. Given line is obviously parallel to the plane \[x-2y+z=6.\]
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