A line passes through the two points `A(2,-3,-1)` and `B(8,-1,2)`. The coordinates of point on this line nearer to the origin and at a distance of 14 units from is

Explanation

Let P be `(x_(1),y_(1),z_(1))`. Point M is `(x_(1),0,z_(1))` and N is `(x_(1),y_(1),z_(1))`.
So normal to plane OMN is `vec(OM)xxvec(ON)=vecx` (say). Therefore,
`|[hati,hatj,hatk],[x_(1),0,z_(1)],[x_(1),y_(1),0]|=hati(-y_(1)z_(1))-hatj(-x_(1)z_(1))+hatk(x_1y_(1))`
`sintheta=(-x_(1)y_(1)z+x_(1)y_(1)z+x_(1)y_(1)z_(1))/(sqrt(x_(1)^(2)+y_(1)^(2)+z_(1)^(2)sqrtsumx_(1)^(2)y_(1)^(2)))`
`("because"sintheta=(vecnxxvec(OP))/(vec|n||vec(P)|))`
`impliescosec^(2)theta=(sumx_(1)^(2)sumx_(1)^(2)y_(1)^(2))/((x_(1)y_(1)z_(1))^(2))=(sumx_(1)^(2))/(x_(1)^(2))+(sumx_(1)^(2))/(y_(1)^(2))+(sumx_(1)^(2))/(z_(1)^(2))`
Now, `sinalpha=(vec(OP).hatk)/(|vec(OP|))=(z_(1))/(sqrtOx_(1)^(2)),sinbeta=(x_(1))/(sqrtOx_(1)^(2))` and `singamma=(y_(1))/sqrt(sumx_(1)^(2))`
Now, `cosec^(2)alpha+cosec^(2)beta+coses^(2)gamma`
`=(sumx_(1)^(2))/(x_(1)^(2))+(sumx_(1)^(2))/(y_(1)^(2))+(sumx_(1)^(2))/(z_(1)^(2))=coses^(2)theta`
Hence proved.