Answer
$3x+y+z=5$
Work Step by Step
When the plane passing through the point $P(a,b,c)$ and normal to vector $n=pi+qj+rk$ then we have a component equation as follows:
$a(x-a)+b(y-b)+c(z-c)=0$ ...(x)
Here, $P(1,-1,3)$ and $n=3i+j+k$
From Equation (x), we get
$3(x-1)+1(y+1)+1(z-3)=0$
$\implies 3x-3+y+1+z-3=0$
Thus, $3x+y+z=5$
