The given equation of lines are
3x – y + 1 = 0 ... (i)
x – 2y + 3 = 0 ... (ii)
Clearly slope of (i) is 3 and slope of (ii) is

1

2

. Since, we have given that both these lines inclined to the same line y = mx + 4, whose slope is m, then we must have
|

m−3

1+3m

| = |

m− 1 2

1+m 1 2

| ⇒ |

m−3

1+3m

| = |

2m−1

2+m

| ⇒

m−3

1+3m

= ± (

2m−1

2+m

)
Now we have two cases
Case (I) :

m−3

1+3m

=

2m−1

2+m

⇒ (m – 3) (2 + m) = (1 + 3m) (2m – 1)
⇒ 2m + m2 – 6 – 3m = 2m – 1 + 6m2 – 3m
⇒ – 5m2 – 5 = 0 ⇒ 5m2 + 5 = 0 ⇒ m2 = – 1, which is not possible.
Case (ii) :

m−3

1+3m

= - (

2m−1

2+m

)
⇒ (m – 3) (2 + m) = – (2m – 1) (1 + 3m)
⇒ 2m + m2 – 6 – 3m = – (2m + 6m2 – 1 – 3m)
⇒ 6m2 – m – 1 + m2 – 6 – m = 0 ⇒ 7m2 – 2m – 7 = 0
⇒ m =

2±(−2)2−4×7×(−7)

2×7

=

2±√4+196

14

=

2±√200

14

=

2±10√2

14

Hence, m =

1±5√2

7