Question

Given point $P(-1,-5,-10)$ and $O$ is the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$, then $PO=$

• A. $13$
• B. $169$
• C. $5$
• D. $12$

Answer 1

The correct option is A $13$

Any general point lying on line

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$ [let]

$\Rightarrow O(3\lambda +2,4\lambda -1,12\lambda +2)$

Since, point $O$ lies on plane $x-y+z=5$.

So, it should also satisfy equation of plane

i.e. $3\lambda +2-4\lambda +1+12\lambda +2=5$

$\Rightarrow \lambda =0$

So, point $O$ is $(2,-1,2)$

On applying distance formula between points

$OP=\sqrt{(3{)}^2+(4{)}^2+(12{)}^2}$

$=\sqrt{9+16+144}$

$=\sqrt{169}=13$