Equation of lines L1:2x−2y+3z−2=0=x−y+z+1=0 L2:x+2y−z−3=0=3x−y+2z−1=0 Distance of point P(0,0,0) from the plane containing L1 and L2 and measured along the line x=y=z is a√ab (where a and b are coprime numbers) then a+b is

Equation of lines L1:2x−2y+3z−2=0=x−y+z+1=0 L2:x+2y−z−3=0=3x

1.	Equation of lines L1:2x−2y+3z−2=0=x−y+z+1=0 L2:x+2y−z−3=0=3x−y+2z−1=0 Distance of point P(0,0,0) from the plane containing L1 and L2 and measured along the line x=y=z is a√ab (where a and b are coprime numbers) then a+b is
Answer» Equation of lines $L_{1} : 2 x - 2 y + 3 z - 2 = 0 = x - y + z + 1 = 0$ $L_{2} : x + 2 y - z - 3 = 0 = 3 x - y + 2 z - 1 = 0$ Distance of point $P (0, 0, 0)$ from the plane containing $L_{1}$ and $L_{2}$ and measured along the line $x = y = z$ is $\frac{a \sqrt{a}}{b}$ (where $a$ and $b$ are coprime numbers) then $a + b$ is