The equation $x^2-3 x y+\lambda y^2+3 x-5 y+2=0$, where $\lambda$ is real number, represents pair of lines. If $\theta$ is acute angle between the lines, then $\frac{\operatorname{cosec}^2 \theta}{\sqrt{10}}=$

Using the condition for a pair of straight lines, the given equation satisfies $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. For $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$, the coefficients are $a=1$, $h=-\frac{3}{2}$, $g=\frac{3}{2}$, $b=\lambda$, $f=-\frac{5}{2}$, $c=2$.

Substituting into the condition:
$2\lambda + 2\left(\frac{-5}{2}\right)\left(\frac{3}{2}\right)\left(\frac{-3}{2}\right) - 1\cdot\left(\frac{25}{4}\right) - \lambda\cdot\left(\frac{9}{4}\right) - 2\cdot\left(\frac{9}{4}\right) = 0$
$2\lambda + \frac{45}{4} - \frac{25}{4} - \frac{9\lambda}{4} - \frac{18}{4} = 0$
$2\lambda - \frac{9\lambda}{4} + \frac{2}{4} = 0$
$\frac{-\lambda + 2}{4} = 0$
$\lambda = 2$

For the acute angle $\theta$ between the lines, $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$. With $a=1$, $b=2$, $h=-\frac{3}{2}$:
$h^2 - ab = \frac{9}{4} - 2 = \frac{1}{4}$
$\tan \theta = \left| \frac{2 \cdot \frac{1}{2}}{3} \right| = \frac{1}{3}$

Evaluating the trigonometric expression, $\frac{\csc^2 \theta}{\sqrt{10}} = \frac{1 + \cot^2 \theta}{\sqrt{10}}$. With $\cot \theta = 3$:
$\csc^2 \theta = 1 + 9 = 10$
$\frac{10}{\sqrt{10}} = \frac{10\sqrt{10}}{10} = \sqrt{10}$

The value is $\boxed{D}$.