Step 1 :
Consider the point (1,-1).
Substituting for x and y in the equations we get,
$d_1 = \bigg| \large\frac{4(1)+3(-1)+10}{\sqrt{4^2+3^2}} \bigg|$
$ = \large\frac{11}{5}$
$d_2 = \bigg| \large\frac{5(1)+(-12)(-1)+26}{\sqrt{25+144}} \bigg|$
$ = \large\frac{43}{13}$
$d_3 = \bigg| \large\frac{7(1)+24(-1)-50}{\sqrt{49+576}} \bigg|$
$ = \large\frac{64}{25}$
Hence $ d_1 \neq d_2 \neq d_3$
Consider the point (1,1)
$ \therefore d_1 = \bigg| \large\frac{4(1)+3(1)+10}{\sqrt{4^2+3^2}} \bigg|$
$ = \large\frac{17}{25}$
$d_2 = \bigg| \large\frac{5(1)-12(1)+26}{\sqrt{25+144}} \bigg|$
$ = \large\frac{19}{13}$
$d_3 = \bigg| \large\frac{7(1)+24(1)-50}{\sqrt{7^2+24^2}} \bigg|$
$ = \large\frac{19}{25}$
$ \therefore d_1 \neq d_2 \neq d_3$
Consider the point (0,0)
$ d_1 = \bigg| \large\frac{4(0)+3(0)+10}{\sqrt{4^2+3^2}} \bigg|$
$ = \large\frac{10}{5}$ = 2 units.
$d_2 = \bigg| \large\frac{5(0)-12(0)+26}{\sqrt{25+144}} \bigg|$
$ = \large\frac{26}{13}$ = 2 units
$d_3 = \bigg| \large\frac{7(0)+24(0)-50}{\sqrt{7^2+24^2}} \bigg|$
$ = \large\frac{50}{25}$ = 2 units
It is clear that $d_1=d_2=d_3$
Hence 'C' is the correct options.