If the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$ meets the plane $x+2 y+3 z=15$ at the point…
If the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$ meets the plane $x+2 y+3 z=15$ at the point $\mathrm{P}$, then the distance of $\mathrm{P}$ from the origin is
$\frac{7}{2}$ units
$\frac{9}{2}$ units
$\frac{\sqrt{5}}{2}$ units
$2 \sqrt{5}$ units
Solution
Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=\mathrm{k}$ (say)
Let $\mathrm{P}$ be the any point on the above line.
$\therefore \quad \mathrm{P}=(2 \mathrm{k}+1,3 \mathrm{k}-1,4 \mathrm{k}+2)$
The point $\mathrm{P}$ lies on the plane
$\begin{array}{ll}
\therefore & 2 \mathrm{k}+1+2(3 \mathrm{k}-1)+3(4 \mathrm{k}+2)=15 \\
\therefore & 2 \mathrm{k}+1+6 \mathrm{k}-2+12 \mathrm{k}+6=15 \\
\therefore & 20 \mathrm{k}=10 \\
\therefore & \mathrm{k}=\frac{1}{2} \\
\therefore & \mathrm{P}=\left(2, \frac{1}{2}, 4\right)
\end{array}$
Distance of $\mathrm{P}$ from origin is
$\sqrt{2^2+\left(\frac{1}{2}\right)^2+4^2}=\sqrt{\frac{81}{4}}=\frac{9}{2}$