A) \[a=2,\,b=8\]
B) \[a=4,b=6\]
C) \[a=6,b=4\]
D) \[a=8,b=2\]
Correct Answer: C
Solution :
[c] Equation of line through \[(5,1,\alpha )\] and \[(3,b,1)\]is \[\frac{x-5}{-2}=\frac{y-1}{b-1}=\frac{z-a}{1-a}=\lambda \] \[\therefore \] Any point on this line is a \[[-2\lambda +5,(b-1)\lambda +1,(1-a)\lambda +a]\] It crosses yz plane where \[-2\lambda 5=0\] \[\lambda =\frac{5}{2}\] \[\therefore \left( 0,(b-1)\frac{5}{2}+1,(1-a)\frac{5}{2}+a \right)=\left( 0,\frac{17}{2}.\frac{-13}{2} \right)\] \[\Rightarrow (b-1)\frac{5}{2}+1=\frac{17}{2}\] and \[(1-a)\frac{5}{2}+a=-\frac{13}{2}\] \[\Rightarrow b=4\,\,and\,\,a=6\]
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