# The distance between Point (-1 ,-5,-10 ) and the point of intersection of Line x-2 / 3=y+1 / 4 =z-2 / 12 with the plane x-y+z=5 .

Solution: We have ,

$L_{1}:x-2/3=y+1/4=z-2/12$ , and   $P:x-y+z=5$

Now ,   $L_{1}:x-2/3=y+1/4=z-2/12=r$

General point on the Given Line $L_{1}$   ,$A(3r+2,4r-1,12r+2)$

$\because$     Point $A$ Lies on thr plane $P$

$\therefore$           $(3r+2)-(4r-1)+(12r+2)=5$  $\Rightarrow$ $r=0$

Point  $A(2,-1,2)$  ,Hence distance between $A(2,-1,2)$ and $B(-1,-5,-10)$

$\Rightarrow$         $AB=\sqrt{9+36+144}=\sqrt{189}$

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