The equation of the plane passing through the point (1, 2, –3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is:

Concept:

The equation plane passing through (x₀, y₀, z₀) and having direction ratios (a, b, c) is given by:

a(x - x0) + b(y - y0) + c(z - z0) = 0

Calculation:

Given, the plane is perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7

∴ Normal vector of required plane is 

\(\vec{n}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1 \end{array}\right|\)

= \(-11 \hat{\imath}-\hat{\jmath}-17 \hat{k}\)

∴ Direction ratios are (- 11, - 1, - 17)

Also, the plane passing through the point (1, 2, –3).

∴ The equation of plane is 11(x - 1) + (y - 2) + 17(z + 3) = 0

⇒ 11x - 11 + y - 2 + 17z + 51 = 0

⇒ 11x + y + 17z + 38 = 0