A) \[2\sqrt{14}\]
B) \[\sqrt{14}\]
C) \[2\sqrt{7}\]
D) \[14\]
Correct Answer: A
Solution :
\[\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=\lambda \] \[x=3\lambda +2,y=2\lambda -1,z=-\lambda +1\] Intersection with plane \[2x+3yz+13=0\] \[2(3\lambda +2)+3(2\lambda -1)-(-\lambda +1)+13=0\] \[13\lambda +13=0\,\] \[\therefore \]\[P(-1,-3,2)\] Intersection with plane \[3x+y+4z=16\] \[3(3\lambda +2)+(2\lambda -1)+(-\lambda +1)=16\] \[\lambda =1\] \[Q\left( 5,1,0 \right)\] \[PQ=\sqrt{{{6}^{2}}+{{4}^{2}}+{{2}^{2}}}=\sqrt{56}=2\sqrt{14}\]You need to login to perform this action.
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