A) \[(1, - 2, 5)\]
B) \[(1, 0, 5)\]
C) \[(0, 3, -5)\]
D) \[(-1, -3, 0)\]
Correct Answer: B
Solution :
Equation of the plane containing the line \[\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\]is \[a(x-1)+b(y-2)+c(z-3)=0\] ..(i) where \[a.1+b.2+c.3=0\] i.e.,\[a+2b+3c=0\] .... (ii) Since the plane (i) parallel to the line \[\frac{x}{1}=\frac{y}{1}=\frac{z}{4}\] \[\therefore \]\[a.1+b.1+c.4=0\]i.e., \[a+b+4c=0\] ... (iii) From (ii) and (iii), \[\frac{a}{8-3}=\frac{b}{3-4}=\frac{c}{1-2}=k\](let) \[\therefore \]\[a=5k,b=-k,c=-k\] On putting the value of a, b and c in equation (i),\[5(x-1)-(y-2)-(z-3)=0\] \[\Rightarrow \]\[5x-y-z=0\] ... (iv) when x = 1, y = 0 and z = 5; then L.H.S. of equation (iv) \[=5x-y-2\] \[=5\times 1-0-5\] \[=0\] = R.H.S. of equation (iv) Hence coordinates of the point (1, 0, 5) satisfy the equation plane represented by equations (iv), Therefore the plane passes through the point (1,0,5)
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