Solve this following


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 a point $P$, then the distance of $P$ from the origin is

  1. $\frac{9}{2}$

  2. $2 \sqrt{5}$

  3. $\frac{\sqrt{5}}{2}$

  4. $\frac{7}{2}$

Correct Option: 1


Any point on the given line can be

$(1+2 \lambda,-1+3 \lambda, 2+4 \lambda) ; \lambda \in \mathrm{R}$

Put in plane

$1+2 \lambda+(-2+6 \lambda)+(6+12 \lambda)=15$

$20 \lambda+5=15$

$20 \lambda=10$


$\therefore$ Point $\left(2, \frac{1}{2}, 4\right)$

Distance from origin




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