The symmetric form of the line of intersection of two planes 3x−y=6,y+2z−1=0

The given planes are $3 x - y = 6, y + 2 z - 1 = 0$

Plug in $z = 0$ to get a point on the line.

$3 x - y = 6, y = 1$

solve the above two equations, we get $x = \frac{7}{3}, y = 1$

A point on the line is $(\frac{7}{3}, 1, 0)$

Vector along the line is cross product of two normal vectors to the planes

it gives $n_{1} \times n_{2} = |\begin{matrix} i & j & k \\ 3 & - 1 & 0 \\ 0 & 1 & 2 \end{matrix}| = i (- 2) - j (6) + k (3)$

Direction ratios of a line of intersection of two planes are proportional to $⟨ 2, 6, - 3 ⟩$

Therefore, the equation of the required line is $\frac{x - \frac{7}{3}}{2} = \frac{y - 1}{6} = \frac{z}{- 3}$

The symmetric form of the line of intersection of two planes $3 x - y = 6, y + 2 z - 1 = 0$