Find the values of non-negative real number `h_1, h_2, h_3, k_1, k_2, k_3` such that the algebraic sum of the perpendiculars drawn from the points `(2,k_1),(3,k_2),*7,k_3),(h_1,4),(h_2,5),(h_3,-3)` on a variable line passing through (2, 1) is zero.

Explanation

Let the equation of variable line be
`ax+by+c=0 " " (1)`
It is given that
`sum_(i=1)^(6)(ax_(i)+by_(i)+c)/(sqrt(a^(2) + b^(2))) = 0, "where" (x_(i), y_(i)), i=1,2,3,4,5,6 " are given points."`
`rArr a((sum_(i=1)^(6)x_(i))/(6)) + b((sum_(i=1)^(6)y_(i))/(6)) +c =0 " " (2)`
Comparing (1) and (2), we get
`x/((sum_(i=1)^(6)x_(i))/(6))=y/((sum_(i=1)^(6)y_(i))/(6))=1`
`therefore (x,y) -= ((sum_(i=1)^(6)x_(i))/(6), (sum_(i=1)^(6)y_(i))/(6))`
This is the fixed point through which the variable (1) passes.
But it is given that the variable line is passing through the point (2,1).
`" So, "(sum_(i=1)^(6)x_(i))/(6) = 2" and " (sum_(i=1)^(6)y_(i))/(6) = 1`
`rArr (2+3+7+h_(1) + h_(2)+h_(3))/(6) =2 " and " (k_(1) + k_(2) + k_(3) +4+5-3)/(6) = 1`
`rArr h_(1) + h_(2)+h_(3) =0 " and " k_(1) + k_(2) + k_(3) = 0`
`" Since " h_(1), h_(2),h_(3), k_(1), k_(2),k_(3) " are non-negative,"h_(1) = h_(2) =h_(3)=0 and k_(1) =k_(2)=k_(3)=0.`