A) 2
B) 3
C) 6
D) 7
Correct Answer: D
Solution :
Let P (3, 8, 2) be the given point. Let the equation of a line through P (3, 8, 2) and parallel to the plane \[3x+2y-2z+15=0\] be \[\frac{x-3}{a}=\frac{y-8}{b}=\frac{z-2}{c}\] ?(i) Then, normal to the plane is perpendicular to the parallel line. Then, \[3a+2b-2c=0\] ?(ii) Line (i) intersects the line \[\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\]at Q. \[\therefore \]\[\left| \begin{matrix} 3-1 & 8-3 & 2-2 \\ a & b & c \\ 2 & 4 & 3 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix} 2 & 5 & 0 \\ a & b & c \\ 2 & 4 & 3 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[2(3b-4c)-5(3a-2c)=0\] \[\Rightarrow \]\[15a-6b-2c=0\] ?(iii) On solving Eqs. (ii) and (iii) by cross-multiplication, we get \[\frac{a}{-4-12}=\frac{b}{-30+6}=\frac{c}{-18-30}\] \[\Rightarrow \]\[\frac{a}{-16}=\frac{b}{-24}=\frac{c}{-48}\Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{6}\] Substituting a, b and c in Eq. (i), we get \[\frac{x-3}{2}=\frac{y-8}{3}=\frac{z-2}{6}=\lambda \][say] Let the coordinates of Q be \[(2\lambda +3,3\lambda +8,6\lambda +2).\]It lies on the line \[\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\] \[\therefore \]\[\lambda +1=\frac{3\lambda +5}{4}=2\lambda \]\[\Rightarrow \]\[\lambda =1\] \[\therefore \]Coordinate of Q =(2 + 3, 3 + 8, 6 + 2) i.e. Q = (5,11,8). Hence, \[PQ=\sqrt{{{(5-3)}^{2}}+{{(11-8)}^{2}}+{{(8-2)}^{2}}}\] \[=\sqrt{4+9+36}=7\]You need to login to perform this action.
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