# Find the equation of the plane passing through the point (-1,3,2) and perpendicular to the planes $x+2y+3z= 5\:\; and\:\; 3x+3y+z= 0\cdot$

Let A,B,C be the d,r's of normal to the required plane. so $A\left ( x+1 \right )+B\left ( y-3 \right )+C\left ( z-2 \right )= 0---(i)$
As (i) is $\perp ^{r}$  to the planes $x+2y+3z= 5\; \; and\; \; 3x+3y+z= 0$   so,
$A+2B+3C= 0---(ii)$
$3A+3B+C= 0---(iii)$
solving (ii) and (iii) we get $\frac{A}{-7}= \frac{B}{8}= \frac{C}{-3}$
ie the d.r's are 7,-8,3
by (i) $\left ( 7x+1 \right )-8\left ( y-3 \right )+3\left ( z-2 \right )= 0$
ie  $7x-8y+3z+25= 0$

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