The equation of the plane is $\left(x+y+z+3\right)+\lambda \left(x-y+z-2\right)=0$
$\Rightarrow x\left(1+\lambda \right)+y\left(1-\lambda \right)+z\left(1+\lambda \right)+3-2\lambda =0$
A point which divides $\left(3,\mathrm{}0,\mathrm{}2\right)$ and $\left(0,\mathrm{}3,\mathrm{}-1\right)$ in $2\mathrm{}:1$ internally is

$\left(\frac{2\times 0+1\times 3}{3},\frac{2\times 3+0\times 1}{3},\frac{2\times \left(-1\right)+1\times 2}{3}\right)$
$\equiv \left(1,\mathrm{}2,\mathrm{}0\right)$
This point must satisfy the equation of the plane

$\Rightarrow 1+\lambda +2-2\lambda +0+3-2\lambda =0$
$\Rightarrow 6-3\lambda =0\Rightarrow \lambda =2$
Equation of the required plane is $3x-y+3z-1=0$
$\Rightarrow 6x-2y+6z=2$
$\Rightarrow a=6,\mathrm{}b=6,\mathrm{}c=2$
Hence, $3a+4b-5c=18+24-10=32$