$AC : \lambda x +(1-\lambda) y +\lambda=0$
Vertex $A$ is on $y$-axis
$\Rightarrow x=0$
So $y =\frac{4}{\lambda}, y =\frac{\lambda}{\lambda-1}$
$\Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1}$
$\Rightarrow \lambda=2$
$AB : 3 x +2 y =4$
$AC : 2 x - y +2=0$
$\Rightarrow A (0,2) \text { Let } C (\alpha, 2 \alpha+2)$
Now (Slope of Altitude through C) $\left(-\frac{3}{2}\right)=-1$
$\left(\frac{2 \alpha}{\alpha-1}\right)\left(-\frac{3}{2}\right)=-1 \Rightarrow \alpha=-\frac{1}{2}$
So $C \left(-\frac{1}{2}, 1\right)$
Let Equation of tangent be $y=m x+\frac{3}{2 m}$
$m ^2+2 m -3=0$
$\Rightarrow m =1,-3$
So tangent which touches in first quadrant at $T$ is
$T \equiv\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right)$
$\equiv\left(\frac{3}{2}, 3\right)$
$\Rightarrow CT =\sqrt{4+4}=2 \sqrt{2}$
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