The line of intersection of the planes →r⋅(3^i−^j+^k)=1 and →r⋅(^i+4^j−^k)=2, is.

→n=→n1×→n2∣∣∣∣∣^i^j^k3−114−1∣∣∣∣∣=^i(−1−(−4))−^j(3−(−1))+^k(3−(−1))→n=7^i+7^j+13^k
Now
3x−y+z=1
x+4y−z=2
Let z=0
3x−y=1
Multiply by 4
12x−4y=4
x+4y=2
Add the equations
13x=6
x=6/13
Substitute to find y
3(6/13)−y=1
18/13−y=1
y=18/13−1=5/13
Therefore
x−6/13=y−5/13=z
or x−132=y−7=z−53