Answer
$3x+y+z=5$
Work Step by Step
The plane passing through the point $P(x_0,y_0,z_0)$ and normal to vector $n=ai+bj+ck$ is represented by a component equation such as:
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0~~~~~$ (1)
Since, we have $P(1,-1,3)$ and $n=3i+j+k$
Thus, from the equation (1), we have
$3(x-1)+1(y+1)+1(z-3)=0$
or, $3x-3+y+1+z-3=0$
Hence, $3x+y+z=5$
