Let P (a, b, g) be the point at which the given line crosses the XZ plane.
Now the equation of given line AB is
Since P \((\alpha, \beta, \gamma)\) lies on line (i)
Also P \((\alpha, \beta, \gamma)\) lie on XZ plane, i.e., y = 0 (0x + 1y + 0z = 0)
Hence, the co-ordinates of required point P is
\(\therefore\) Co-ordinate of required point is \((\frac{17}{3},0,\frac{23}{3})\)
Let q be the angle made by line AB with XZ plane.