The symmetric form of the line of intersection of planes $3x+2y+z-5=0$ and $x+y-2z=0$

Solution:

The given planes are $x+y-2z=0$ and $3x+2y+z-5=0$ $x=5,y=-5$

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

$x+y=0,3x+2y-5=0$

solve the above two equations, we get $x=5,y=-5$ 

A point on the line is $\left(5,-5,0\right)$

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

it gives $n_1\times n_2=\left|\begin{array}{ccc}i& j& k\\ 1& 1& -2\\ 3& 2& 1\end{array}\right|=i\left(5\right)-j\left(7\right)+k(-1)$

Direction ratios of a line of intersection of two planes are proportional to $⟨5,-7,-1⟩$

Therefore, the equation of the required line is $\boxed{\frac{x-5}{5}=\frac{y+5}{-7}=\frac{z}{-1}}$