The equations of the lines through the origin making an angle of \[{60^\circ }\] with the line \[x + y\sqrt 3 + 3\sqrt 3 = 0\] are
A. \[y = 0,x - y\sqrt 3 = 0\]
В. \[x = 0,x - y\sqrt 3 = 0\]
C. \[x = 0,x + y\sqrt 3 = 0\]
D. \[y = 0,x + y\sqrt 3 = 0\]

Hint: In general, the equation y = mx indicates a straight line with gradient m that passes through the origin. \[y = mx\] is the equation for a straight line with gradient m going through the origin.

Formula Used:
A line's slope intercept is given as \[y = mx\] Where, the variable \[m\] in the formula is the slope.

Complete step by step Solution:
The line through the origin is said by the equation as below
\[y = mx\] --- (1)
We have been given in the question that the equation of the line is
 \[x + y\sqrt 3 + 3\sqrt 3 = 0\] ---- (2)
The above line has the slope as
\[ - \dfrac{1}{{\sqrt 3 }}\]
We have been given in the question that the angle between (1) and (2) is \[{60^\circ }\]
Then we have
\[\left| {\dfrac{{m + \dfrac{1}{{\sqrt 3 }}}}{{1 - \dfrac{m}{{\sqrt 3 }}}}} \right| = \tan {60^\circ }\]
Now, we have to determine the value of \[\tan {60^\circ }\]in the above equation and also solve the numerator and denominator, we get
\[\left| {\dfrac{{\sqrt 3 \;{\rm{m}} + 1}}{{\sqrt 3 - {\rm{m}}}}} \right| = \sqrt 3 \]
On removing the modulus from the above equation, we get
\[\dfrac{{\sqrt 3 \;{\rm{m}} + 1}}{{\sqrt 3 - {\rm{m}}}} = \pm \sqrt 3 \]
Now, let’s move the term \[\sqrt 3 - 1\] from denominator of the left hand side to the right hand side of the equation, we get
\[\sqrt 3 \;{\rm{m}} - 1 = \pm \sqrt 3 (\sqrt 3 - {\rm{m}})\]
On multiplying the term \[ \pm \sqrt 3 \]with the terms inside the parentheses, we get
\[\sqrt 3 \;{\rm{m}} - 1 = 3 - \sqrt 3 \;{\rm{m}}\]
Now, we have to solve for \[m\], we get
\[{\rm{m}} = \dfrac{1}{{\sqrt 3 }}\]
Therefore, the equations of the lines through the origin making an angle of \[{60^\circ }\] with the line \[x + y\sqrt 3 + 3\sqrt 3 = 0\] are \[y = \dfrac{x}{{\sqrt 3 }}\] or \[x - y\sqrt 3 \]

Therefore, the correct option is (B).

Note:Any straight line, such as \[ax + by + c = 0\] that goes through the origin will satisfies the point (0,0) when (0,0) is introduced into the line.
That is,\[a\left( 0 \right) + b\left( 0 \right) + c = 0\]
\[ \Rightarrow C = 0\]
As a result, the value of C will be zero.