A) \[(3d,\text{ }2a,\text{ }3a),\text{ }(a,\text{ }a,2a)\]
B) \[(3a,\text{ }2a,\text{ }3a),\text{ }(a,\text{ }a,\,a)\]
C) \[(3a,\text{ }3a,\text{ }3a),\text{ (}a,\text{ }a,\text{ }a\text{)}\]
D) \[(2a,\text{ }3a,\text{ }3a),\text{ (2}a,\text{ }a,\text{ }a\text{)}\]
Correct Answer: B
Solution :
Equation of given lines can be written in .symmetrical form \[\frac{x}{1}=\frac{y+a}{1}=\frac{z}{1}=r\] and \[\frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}=\lambda \] Any point on the line is \[P=(r,\text{ }r-a,\text{ }r)\] and\[Q=(2\lambda ,\text{ }-a,\,\,\lambda ,\,\lambda )\] DRs of PQ are\[r-2\lambda +a,r-\lambda -a,r-\lambda \]. According to the given condition, \[\frac{r-2\lambda +a}{2}=\frac{r-\lambda -a}{1}=\frac{r-\lambda }{2}\] Taking pairwise 1st and IIIrd, we get \[\lambda =a\] and Und and IIIrd, we get \[r-\lambda =2a\] \[\therefore \] \[r=3a,\lambda =a\] \[\therefore \] \[P=(3a,\text{ }2a,\text{ }3a)\] and\[Q=(a,\text{ }a,\text{ }a)\]
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