Eqn. of xy plane is z = 0
the eqn. of line joining the points P(1, -2, 3) and Q(4, 7, 8) is
\(\frac{x-1}{3}=\frac{y+2}{9}=\frac{z-3}{5}\)
any point on this line is \((1+3 \lambda,-2+9 \lambda, 3+5 \lambda)\)
this lies on xy plane, \(3 + 5 \lambda=0,5 \lambda=-3, \lambda=-\frac{3}{5}\)
Required point of intersection of plane and line is
\( \left(1+3\left(-\frac{3}{5}\right),-2+9 \cdot\left(-\frac{3}{5}\right), 3+5\left(-\frac{3}{5}\right)\right) \)
\(=\left(1-\frac{2}{5},-2-\frac{27}{5}, 0\right)=\left(-\frac{4}{5}, \frac{-37}{5}, 0\right)\)
