Solution The equation of the plane containing the given point is
A (x – 1) + B(y + 1) + C (z – 2) = 0
Applying the condition of perpendicularly to the plane given in (1) with the planes (1)
2x + 3y – 2z = 5 and x + 2y – 3z = 8, we have
2A + 3B – 2C = 0 and A + 2B – 3C = 0
Solving these equations, we find A = – 5C and B = 4C Hence, the required
equation is
– 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0
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