Questions
a
1
b
2
c
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d
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detailed solution
Correct option is A
Equation of line through A(1,−2,3) and parallel to the line x2=y3=z−1−6 is x−12=y+23=z−3−6=r………(1) Any point P on this line is P(2r+1,3r−2,−6r+3) . If P is the point of intersection of line (1) and plane x−y+z−5=0, then 2r+1−3r+2−6r+3−5=0⇒r=17∴P97,−117,157 Required distance =AP=1−972+−2+1172+3−1572=449+949+3649=1Talk to our academic expert!
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