Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

$P_1:4x+6y-7z=2$ $P_2:8x+12y-14z=2\Rightarrow 4x+6y-7z=1$ Distance between $P_1$ and $P_2$ $\begin{array}{rl}& =\left|\frac{d_2-d_1}{\sqrt{a_1{a}_2+b_1{b}_2+c_1{c}_2}}\right|\\ & =\left|\frac{2-1}{\sqrt{4\times 4+6\times 6+(-7)\times (-7)}}\right|\\ & =\frac{1}{\sqrt{16+36+49}}=\frac{1}{\sqrt{101}}\end{array}$ By comparing with $\frac{1}{\sqrt{N}},N=101$ Therefore, $\frac{N(N+1)}{2}=\frac{101\times 102}{2}=5151$