Find the equation of the plane through and perpendicular to both planes in Problem 22.

The given point is $(-4,-1,2),$ and the planes are$2x-y-z=4$and$3x-2y-6z=7.$

The equation of the plane if $N=ai+bj+ck$is perpendicular to the plane and passing through $(x_0,y_0,z_0)$is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0,$
where $(x,y,z)$is any other point in the plane.

Consider the planes $2x-y-z=4$and $3x-2y-6z=7.$The normal vectors to these planes are $2i-j-k$and $3i-2j-ck.$The cross product of two vectors is perpendicular to the plane, that is,

$=N\left|\begin{array}{ccc}i& j& k\\ 2& -1& -1\\ 3& -2& -6\end{array}\right|\begin{array}{}\\ \\ \end{array}$

$$\begin{gathered} =i(6-2)-j(-12+3)+k(-4+3) \\ =4i+9j-k \end{gathered}$$

It is known that, the equation of the plane is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0.$

Here, $N=4i+9j-k$and $(x_0,y_0,z0)=(-4,-1,2).$Substitute the values in $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$and simplify.

$$\begin{gathered} 4(x+4)+9(y+1)-1(z-2)=0 \\ \Rightarrow 4x+16+9y+9-z+2=0 \\ \Rightarrow 4x+9y-z+27=0 \end{gathered}$$

Hence, the equation of the plane is$4x+9y-z+27=0$