The equations of the sides AB and AC of a triangle ABC are  $(\lambda +1)x+\lambda y=4$ and  $\lambda x+(1-\lambda )y+\lambda =0$ respectively. Its vertex A is on the y-axis and its orthocentre is  $(1,\mathbf{2}).$ The length of the tangent from the point C to the part of the parabola  $y^2=6x$ in the first quadrant is

Let  $A=(0,y_1)$

$\lambda y_1=\mathrm{4........}\left(1\right)$

And  $(1-\lambda )y_1=-\lambda \mathrm{......}\left(2\right)$

From (1) & (2)
$y_1=2$ , A = (0, 2),   $\lambda =2$
$\overline{AB}=3x+2y=4,\overline{AC}=2x-y+2=0$ 
Let  $C=[h,2h+2]$
    $m_1\cdot m_2=-1$     $\Rightarrow \left(\frac{-3}{2}\right)\left(\frac{2h}{h-1}\right)=-1 \Rightarrow h=\frac{-1}{2}\Rightarrow C=(-\frac{1}{2},1)$ 
Equation of tangent is $y=mx+\frac{a}{m} \Rightarrow 1=m\left(\frac{-1}{2}\right)+\frac{3}{2m}$  $\Rightarrow m = 1, -3$
$$\begin{gathered} C=(\frac{-1}{2},1) \\ P=[\frac{a}{m^2},\frac{2a}{m}]=(\frac{3}{2},3) \\ CP=2\sqrt{2} \end{gathered}$$