Solve this following


The distance of the point $\mathrm{P}(3,4,4)$ from the point of intersection of the line joining the points. $\mathrm{Q}(3,-4,-5)$ and $\mathrm{R}(2,-3,1)$ and the plane $2 x+y+z=7$, is equal to




$\Rightarrow(x, y, z) \equiv(r+3,-r-4,-6 r-5)$

Now, satisfying it in the given plane.

We get $r=-2$.

so, required point of intersection is $\mathrm{T}(1,-2,7)$.

Hence, $\mathrm{PT}=7 .$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now