The a1x+b1y+c1 =0 and a2x+b2y+c2 = 0 are || if,

a1/a2 = b1/b2

Here the || lines are x+y-1=0 and 3x-ty-2 = 0.

a1 = 1, b1 =1 . a2 = 3 and b2 = -t. Therefore,

1/3 = 1/(-t) . Multiply by 3t.

t = -3

If the 2 given lines are parallel, then the values of their slopes are equal.

The given equations are x+y-1=0 and 3x-ty-2=0, so we'll have to put them in the standard from, which is y = mx+n.

We'll start with x+y-1=0.

We'll isolate y to the left side:

y = -x+1

So, the slope can be easily determined as m1 = -1.

That means that the second slope has the same value, m2 = -1.

We'll put the other equation into the standard form, isolating y to the left side:

3x-ty-2=0

-ty = -3x + 2

We'll divide by -t:

y = 3x/t - 2/t

The slope m2 = 3/t

But m2 = -1, so 3/t = -1

**t = -3**

**The line d2 is: 3x+3y-2=0**