If A $(1,0,2), \mathrm{B}(2,1,0), \mathrm{C}(2,-5,3), \mathrm{D}(0,3,2)$ are four points and the point of…
If A $(1,0,2), \mathrm{B}(2,1,0), \mathrm{C}(2,-5,3), \mathrm{D}(0,3,2)$ are four points and the point of intersection of the lines AB and CD is $\mathrm{P}(a, b, c)$, then $a+b+c=$
$3$
$-5$
$5$
$-3$
Solution
$\mathrm{A}(1,0,2), \mathrm{B}(2,1,0), \mathrm{C}(2,-5,3), \mathrm{D}(0,3,2)$
Eq. of line AB is $\frac{x-1}{1}=\frac{y}{1}=\frac{z-2}{-2}=\lambda$ ....(i)
Eq. of line CD is $\frac{x-2}{-2}=\frac{y+5}{8}=\frac{z-3}{-1}=\mu$ ....(ii)
$\Rightarrow x=\lambda+1, y=\lambda, z=-2 \lambda+2$ [From (i)]
and $x=-2 \mu+2, y=8 \mu-5, z=-\mu+3$ [From (ii)]
Solving for $\mu$ and $\lambda$ we get $\mu=\frac{3}{5}, \lambda=\frac{7}{5}$
$\therefore x=\frac{12}{5}, y=\frac{7}{5}, z=\frac{-4}{5} \Rightarrow \mathrm{P}(a, b, c)=\left(\frac{12}{5}, \frac{7}{5}, \frac{-4}{5}\right)$
$\Rightarrow a+b+c=3$