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Q.
If the line, $\frac{x-1}{2} =\frac{y+1}{3} =\frac{z-2}{4}$ meets the plane, $x+2y+3z=15$ at a point P, then the distance of P from the origin is
Three Dimensional Geometry
Solution:
Let point on line be $P (2k + 1, 3k - 1, 4k + 2) $
Since, point $P$ lies on the plane $x + 2y + 3z = 15 $
$\therefore 2k + 1 + 6 k - 2 + 12k + 6 = 15$
$\Rightarrow k=\frac{1}{2}$
$ \therefore P \equiv\left(2, \frac{1}{2}, 4\right)$
Then the distance of the point P from the origin is
$ OP=\sqrt{4+\frac{1}{4}+16}=\frac{9}{2}$