Finding the equation of a plane?

[engineer_in_training]

The plane contains point P(−3,1,3) and the line x(t)=2−3ty(t)=2−3tz(t)=−3+4t.

I made two vectors(<-3,-3,4> and <0,-4,1>) and did their cross product, getting <13,3,12> for the normal vector. I then plugged that and the point into the formula for a plane getting 13(x+3)+3(y-1)+12(z-3)=0. But thats wrong. Any idea where I made the error?

 

[pka]

The plane contains point P(−3,1,3) and the line x(t)=2−3ty(t)=2−3tz(t)=−3+4t.

I made two vectors(<-3,-3,4> and <0,-4,1>) and did their cross product, getting <13,3,12> for the normal vector. I then plugged that and the point into the formula for a plane getting 13(x+3)+3(y-1)+12(z-3)=0. But thats wrong. Any idea where I made the error?

I cannot see where \(\displaystyle <0,4,1>\) comes from.
But \(\displaystyle Q2,2,-3)\) is a point on the line.
So the normal you need is \(\displaystyle \overrightarrow {PQ} \times <-3,-3,4>\)