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Q.
The line passing through the points $(5,1, a)$ and $(3, b, 1)$ crosses yz plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$, then
Vector Algebra
Solution:
Equation of line will be
$\frac{x-5}{2}=\frac{y-1}{1-b}=\frac{z-a}{a-1}$
$\Theta $ Point of intersection with $y z$-plane $x=0$
$ \frac{-5}{2}=\frac{y-1}{1-b}=\frac{z-a}{a-1} $
$\therefore \frac{-5}{2}(1-b)+1=\frac{17}{2} \Rightarrow 5 b-3=17 \Rightarrow b=4 $
$\text { and } \frac{-5}{2}(a-1)+a=\frac{-13}{2} \Rightarrow-3 a+5=-13 \Rightarrow a=6$