Let $\mathrm{\alpha x}+\mathrm{\beta y}+yz=1$be the equation of a plane passing through the point (3, -2, 5) and perpendicular to the line joining the points (1, 2, 3) and (-2, 3, 5). Then the value of $\mathrm{\alpha \beta }y$ is equal to _____.

Given question has to be $\mathrm{\alpha x}+\mathrm{\beta y}+\mathrm{\gamma z}=1$, then only we can find  $\text{'}\mathrm{\gamma }\text{'}$
But given $\mathrm{\alpha x}+\mathrm{\beta y}+yz=1$, which is wrong.
If the coefficient of z is $\mathrm{\gamma }$ then 
Dr’s of the line joining the pts $\mathrm{A}(1,2,3),\mathrm{B}(-2,3,5)\text{ is }\pm (3,-1,-2)$
$=(3,-1,-2)\text{ or }(-3,1,2)$
So equation of the plane passing through (3, -2, 5) and Dr’s of the normal to plane (3, -1, -2) is
$\begin{array}{l}\mathrm{a}\left(\mathrm{x}-{\mathrm{x}}_1\right)+\mathrm{b}\left(\mathrm{y}-{\mathrm{y}}_1\right)+\mathrm{c}\left(\mathrm{z}-{\mathrm{z}}_1\right)=0\\ \Rightarrow 3(\mathrm{x}-3)-1(\mathrm{y}+2)-2(\mathrm{z}-5)=0\\ \Rightarrow 3\mathrm{x}-\mathrm{y}-2\mathrm{z}-9-2+10=0\\ \Rightarrow 3\mathrm{x}-\mathrm{y}-2\mathrm{z}-1=0\\ \Rightarrow 3\mathrm{x}-\mathrm{y}-2\mathrm{z}=1\dots \dots ..\left(1\right)\end{array}$
So comparing (1) with given, $\mathrm{\alpha x}+\mathrm{\beta y}+\mathrm{\gamma z}=1$, we get
$\begin{array}{r}\mathrm{\alpha }=3,\mathrm{\beta }=-1,\mathrm{\gamma }=-2\\ \therefore \mathrm{\alpha \beta \gamma }=3(-1)(-2)=6\end{array}$