If \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} - {{\lambda }} = 0\] represents the asymptotes of \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} = 0\], then find the value of${{\lambda }}$.
${\text{A}}$. $3$
${\text{B}}$. $ - 6$
${\text{C}}$. $8$
${\text{D}}$. $12$

Hint: From the given question, we have to find the value of ${{\lambda }}$ and choose the correct answer. First, we have to find the joint equation of asymptotes by the given equation then we have to compare the joint equation to the given asymptotic equation, we get the required result.

Complete step-by-step solution:
We have to find the value of ${{\lambda }}$ by the given equation \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} = 0\].
First, we are going to find the joint equation of asymptotes.
Given that, \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} = 0\].
Now, add and subtract the term $12$ on the left hand side (LHS) of the above term. Then, we get
\[ \Rightarrow {\text{xy}} - 4{\text{x}} + 3{\text{y}} - 12 + 12 = 0\]
Let us separate the common term on the left hand side (LHS)
\[ \Rightarrow {\text{x}}\left( {{\text{y}} - 4} \right) + 3\left( {{\text{y}} - 4} \right) + 12 = 0\]
\[ \Rightarrow {\text{x}}\left( {{\text{y}} - 4} \right) + 3\left( {{\text{y}} - 4} \right) = - 12\]
\[ \Rightarrow \left( {x + 3} \right)\left( {{\text{y}} - 4} \right) = - 12\]
Thus \[\left( {x + 3} \right)\left( {{\text{y}} - 4} \right) = 0\] is the joint equation of asymptotes.
Therefore, \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} - 12 = 0\] represents the asymptotes of \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} = 0\].
Now, for finding the value of ${{\lambda }}$. We have to compare the given asymptotic equation\[{\text{xy}} - 4{\text{x}} + 3{\text{y}} - {{\lambda }} = 0\] with new asymptotic equation \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} - 12 = 0\]. Then we get the desired solution.
Compare \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} - {{\lambda }} = 0\] with \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} - 12 = 0\].
Thus, we get the value of ${{\lambda }} = 12$.

$\therefore $ The correct option is ${\text{D}}$.

Note: We have to know that an asymptote to a curve is the tangent to the curve such that the point of contact is at infinity. In particular the asymptote touches the curve at $ + \infty $ and $ - \infty $.
The equation of the asymptote to the hyperbola $\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1$ .