Q. 84.0( 4 Votes )

Find the equation

Answer :

We know that equation of plane passing through the line of intersection of planes


(a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0


So equation of plane passing through the line of intersection of planes


x – 3y + 2z – 5 = 0 and 2x – y + 3z – 1 = 0 is given by


(x – 3y + 2z – 5) + k(2x – y + 3z – 1) = 0


x(1 – 2k) + y( – 3 – k) + z(2 + 3k) – 5 – k = 0 …… (1)


plane (1) is passing the through the point(1, – 2, 3) so,


1(1 + 2k) + ( – 2)( – 3 – k) + (3)(2 + 3k) – 5 – k = 0


1 + 2k + 6 + 2k + 6 + 9k – 5 – k = 0


8 + 12k = 0


12k = – 8




Put value of k in eq.(1),


x(1 + 2k) + y( – 3 – k) + z(2 + 3k) – 5 – k = 0




Multiplying by( – 13),


x + 7y + 13 = 0




Equation of required plane is,


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