Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given $\mathrm{A}(1,1,\lambda )$ and $\mathrm{B}(-3,0,1)$ are equidistant from $3\mathrm{x}+4\mathrm{y}-12\mathrm{z}+13=0$ $$\begin{array}{rl}& \therefore \left|\frac{3(1)+4(1)-12\lambda +13}{\sqrt{9+16+144}}\right|\\ & \therefore \left|\frac{20-12\lambda }{13}\right|=\left|\frac{-8}{13}\right|\\ \therefore & 20-12\lambda =\pm 8\\ \therefore & 20-12\lambda =8\text{ or }20-12\lambda =-8\Rightarrow \lambda =1\text{ or }\lambda =\frac{7}{3}\end{array}$$