Hello,

The line passes through point ( 1, -2, 3 )

Let the directions are ( a, b, c )

Hence, the equation of line is,

( x - 1 / a ) = ( y - ( -2) / b ) = ( z - 3 / c )

So, ( x - 1 / a ) = ( y + 2 / b ) = ( z - 3 / c )................ ( 1 )

The lines are parallel to planes x - y + 2z = 5 and 3x + y + z = 6

Since, the directions are a, b and c,

We can write, a1.a2 + b1.b2 + c1.c2 = 0

Hence, for first plane,

a - b + 2c = 0........... ( 2 )

For second plane,

3a + b + c = 0........... ( 3 )

Solving eqns ( 2 ) and ( 3 ) by Cramer's rule,

( a / -1-2 )= -( b / 1-6 ) = ( c / 1- (-3))

So, ( a / -3 ) = ( b / 5 ) = ( c / 4 )

Let the constant of proportionality be 1

Hence, a = -3 , b = 5 and c = 4

So, from equation ( 1 ), we get,

( x -1 / -3 ) = ( y + 2 / 5 ) = ( z - 3 / 4 )

It is the required equation of line.

Best Wishes.