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  4. The equation of the plane through the point (2, 5, - 3) perpendicular to the planes x + 2y + 2z = 1 and x - 2y + 3z = 4 is

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Q. The equation of the plane through the point $(2, 5, - 3)$ perpendicular to the planes $x + 2y + 2z = 1$ and $x - 2y + 3z = 4$ is

Three Dimensional Geometry

Solution:

Equation of plane passing through $(2,5, -3)$ is
$a(x - 2) + b(y - 5) + c(z + 3) = 0\quad \ldots (1)$
which is perpendicular to
$x + 2y + 2z = 1$ and $x - 2y + 3z = 4$
then $a + 2b + 2c = 0\quad \ldots (2)$
and $a - 2b + 3c = 0 \quad \ldots (3)$
On solving $(1)$, $(2)$ and $(3)$, we get
$\begin{vmatrix}x-2&y-5&z+3\\ 1&2&2\\ 1&-2&3\end{vmatrix}=0$
$\Rightarrow 10x-y - 4z - 27 = 0$