Direction Ratio's of the line $L_1$

$\left|\begin{array}{ccc}i& j& k\\ 2& -2& 3\\ 1& -1& 1\end{array}\right|$

$=i\left(1\right)-j \left(-1\right)+k\left(0\right)$

$=\left(1, 1, 0\right)$

At $y=0$

$\left.\begin{array}{c}2x+3z=2\\ x+z=-1\end{array}\right]\to$ $(-5,0,4)$ is point

Equation of $L_1$

$\frac{x+5}{1}=\frac{y}{1}=\frac{z-4}{0}=\alpha$

Similarly for $L_2$

$\frac{x-\frac{7}{5}}{3}=\frac{y}{-5}=\frac{z+\frac{8}{5}}{-7}=\beta$

Point of intersection of $L_1$ and $L_2$

$(-1,4,4)$

Direction Ratio's of normal to plane passing through $L_1$ and $L_2$ $=\left|\begin{array}{ccc}i& j& k\\ 1& 1& 0\\ 3& -5& -7\end{array}\right|=i\left(-7\right)-j\left(-7\right)+k\left(-5-3\right)=-7i+7j-8k=\left(-7, 7, -8\right)$

Equation of plane

$-7x+7y-8z=c$ .

$c=3$   (Since plane passes through$(-1,4,4)$ )

Equation of plane

$7x-7y+8 z+3=0$

Distance from origin

$=\frac{\left|3\right|}{\sqrt{7^2+{\left(-7\right)}^2+8^2}}=\frac{3}{\sqrt{162}}=\frac{3}{9\sqrt{2}}$

$=\frac{1}{3\sqrt{2}}$