Chọn: Đáp án B

Ta có:$SB=\sqrt{S{{H}^{2}}+H{{B}^{2}}}=\frac{\sqrt{6}}{2}a$

$H{{C}^{2}}=A{{C}^{2}}+A{{H}^{2}}={{a}^{2}}+\frac{3{{a}^{2}}}{4}=\frac{7{{a}^{2}}}{4}$

$SC=\sqrt{S{{H}^{2}}+H{{C}^{2}}}=\sqrt{\frac{3{{a}^{2}}}{4}+\frac{2{{a}^{2}}}{4}}=\frac{\sqrt{10}}{2}a$

${{S}_{SBC}}=\frac{1}{2}SB.SC=\frac{1}{2}.\frac{\sqrt{6}}{2}a.\frac{\sqrt{10}}{2}a=\frac{\sqrt{15}}{4}{{a}^{2}}$

Vậy $d\left( A;\left( SBC \right) \right)=S\frac{3{{V}_{S.ABC}}}{{{V}_{SBC}}}=\frac{\frac{3}{4}{{a}^{3}}}{\frac{\sqrt{15}}{4}{{a}^{2}}}=\frac{3}{\sqrt{15}}a$