# The line through the points $(5,1,a)\:\:and\:\:(3,b,1)$ crosses $yz$ plane at the point $(0,\large\frac{17}{2},-\frac{13}{2})$, then $(a,b)=?$

$\begin{array}{1 1}(a)\:\:(8,2)\\ (b)\:\:(2,8) \\ (c)\:\:(4,6) \\ (d)\:\:(6,4) \end{array}$

Let $P(5,1,a)\:\:and\:\:Q(3,b,1)$ be the given points.
$\therefore\:$Eqn. of the line $\overline {PQ}$ is $\large\frac{x-3}{2}=\frac{y-b}{1-b}=\frac{z-1}{a-1}$$=\lambda This line crosses yz plane at (0,\large\frac{17}{2},-\frac{13}{2}) \therefore\:(0,\large\frac{17}{2},-\frac{13}{2})$$=(2\lambda+3,\:(1-b)\lambda+b,\:(a-1)\lambda+1)$
$\Rightarrow\:2\lambda+3=0,\:\:\frac{17}{2}=$$(1-b)\lambda+b$ and $(a-1)\lambda+1=-\large\frac{13}{2}$
$\Rightarrow\:\lambda=-\large\frac{3}{2}$
$\large\frac{17}{2}=\frac{-3+5b}{2}\:\:and\:\:-\frac{13}{2}=\frac{-3a+5}{2}$
$\Rightarrow\:b=4\:\:and\:\:a=6$ $i.e., (a,b)=(6,4)$