A) \[\frac{x+4}{3}\,\,\,=\,\,\,\frac{y-3}{-1}\,\,=\,\,\frac{z-1}{1}\]
B) \[\frac{x+4}{1}\,\,\,=\,\,\,\frac{y-3}{1}\,\,=\,\,\frac{z-1}{3}\]
C) \[\frac{x+4}{-1}\,\,\,=\,\,\,\frac{y-3}{1}\,\,=\,\,\frac{z-1}{1}\]
D) \[\frac{x-4}{2}\,\,\,=\,\,\,\frac{y-3}{1}\,\,=\,\,\frac{z+1}{4}\]
Correct Answer: A
Solution :
The equation of required line passing through \[\left( -4,\text{ }3,\text{ }1 \right)\] is \[\frac{x+4}{a}\,\,=\,\,\frac{y-3}{b}\,\,=\,\,\frac{z-1}{c}\] given line is \[\frac{x+1}{-3}\,\,=\,\,\frac{y-3}{2}\,\,=\,\,\frac{z-2}{-1}\] Now the required line is parallel to plane x \[+\,2y-z-5=0\] \[\therefore \] given lines are coplanar \[\because \,\,\,\,\,\left| \begin{align} & -3\,\,\,\,\,\,0\,\,\,\,\,\,\,\,1 \\ & a\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,c \\ & -3\,\,\,\,\,\,2\,\,\,\,\,-1 \\ \end{align} \right|\,\,\,\,\,\,\,=\,\,\,\,\,\,0\] \[3\left( -b+2c \right)+0+2a+3b=0\] \[-2a\,\,+\,\,6c\,\,=\,\,0\] \[\]??. (1) Also \[a+2b-c=0\] \[3c+2b-c=0\] [use (1)] \[2c\,\,+\,\,2b\,\,=\,\,0\] \[b\,\,=\,\,-c\] \[\therefore \] Required line is \[\frac{x+4}{(3c)}\,\,=\,\,\frac{y-3}{-c}\,\,=\,\,\frac{z-1}{c}\] \[\frac{x+4}{3}\,\,=\,\,\frac{y-3}{-1}\,\,=\,\,\frac{z-1}{1}\]
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