Step 1 :

Let (h, k) be the coordinate of the foot of the perpendicular from the given point (2,3). Equation of the given line is $y=3x+4$.

Then the slope of the perpendicular line is $ \large\frac{k-3}{h-2}$

Slope of the line $3x-y+4=0$ is $ -\bigg( \large\frac{3}{-1} \bigg)$$=3$

Hint : Condition for the perpendicularity of lines is $m_1m_2=-1$

$ \therefore \bigg( \large\frac{k-3}{h-2} \bigg)$$ (3)=-1$

$ \Rightarrow (3k-9)=-1(h-2)$

$ \Rightarrow 3k-9=-h+2$

$ \therefore 3k+h=11$---------(1)

Also (h,k) lies on the line $3x-y+4=0$.

$ \therefore 3(h) = k + 4 = 0$

$ \Rightarrow 3h-k=-4$---------(2)

Solving equation (1) and (2) we get,

$ 3k+h=11$

$3h-k=-4$

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$9k+\not{3}\not{h}=33$

$\not{3}\not{h} - k = -4$

$(-) \quad (+) \quad (+)$

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$ \qquad 10k=37$

$ \Rightarrow k = \large\frac{37}{10}$

Substituting this value of k in equation (1) we get,

$ 3 \bigg( \large\frac{37}{10} \bigg)$$+h=11$

$ \Rightarrow h = 11-\large\frac{111}{10}$

$ = -\large\frac{1}{10}$

Hence the coordinate are $ \bigg(- \large\frac{1}{10}$$, \large\frac{37}{10} \bigg)$

Hence B is the correct option.