Câu 3:
\(\begin{array}{l}A\left( { - 1;4} \right) \in \left( C \right) \Leftrightarrow 4 = \frac{{2.1 - 6m + 4}}{{ - m + 2}}\\ \Leftrightarrow - 4m + 8 = 6 - 6m \Leftrightarrow 2m = - 2 \Leftrightarrow m = - 1\end{array}\)
Câu 4:
Gọi M là trung điểm BC
\(\begin{array}{l} \Rightarrow AM = 3OM \Rightarrow d\left( {A,\left( {SBC} \right)} \right) = 3d\left( {O,\left( {SBC} \right)} \right)\\ \Rightarrow {d_1} = 3{d_2} \Rightarrow {d_2} = \frac{1}{3}{d_1} \Rightarrow d = {d_1} + {d_2} = \frac{4}{3}{d_1}\end{array}\)
\(\begin{array}{l}{S_{ABC}} = \frac{{{a^2}\sqrt 3 }}{4},SO = \sqrt {S{A^2} - A{O^2}} = \sqrt {3{a^2} - \frac{{{a^2}}}{3}} = \frac{{2a\sqrt 6 }}{3}\\ \Rightarrow {V_{S.ABC}} = \frac{1}{3}SO.{S_{ABC}} = \frac{1}{3}.\frac{{2a\sqrt 6 }}{3}.\frac{{{a^2}\sqrt 3 }}{4} = \frac{{{a^3}\sqrt 2 }}{6}\\SM = \sqrt {S{B^2} - B{M^2}} = \sqrt {3{a^2} - \frac{{{a^2}}}{4}} = \frac{{a\sqrt {11} }}{2}\\ \Rightarrow {S_{SBC}} = \frac{1}{2}SM.BC = \frac{1}{2}.\frac{{a\sqrt {11} }}{2}.a = \frac{{{a^2}\sqrt {11} }}{4}\\{d_1} = \frac{{3{V_{S.ABC}}}}{{{S_{SBC}}}} = \frac{{3.\frac{{{a^3}\sqrt 2 }}{6}}}{{\frac{{{a^2}\sqrt {11} }}{4}}} = \frac{{2a\sqrt {22} }}{{11}} \Rightarrow d = \frac{4}{3}{d_1} = \frac{4}{3}.\frac{{2a\sqrt {22} }}{{11}} = \frac{{8a\sqrt {22} }}{{33}}\end{array}\)
