Answer:
Let
a, b, c are direction ratios of require plane. Therefore, required equation of
plane passing through(-1, 3, 2) is given by
a(x
+ 1) + b (y ? 3) + c(z ? 2) = 0 ? (1)
Also
direction ratios of plane x + 2y + 3z = 5 are 1, 2, 3 and direction ratios of
plane 3x + 3y + z = 0 are 3, 3, 1
Now
it is given that required plane (1) is perependicular to planes x + 2y + 3z = 5
and 3x + 3y + z = 0
a × 1 + b × 2 +
c × 3 = 0
and
a × 3 + b × 3 + c × 1 = 0
i.e.
a + 2b + 3c = 0 ?. (2)
and
3a + 3b + c = 0 ?.. (3)
On
cross-multiplying (2) and (3), we get
Putting
these values in (1), we get
?7K
(x + 1) + 8K (y ? 3)P ? 3K (z ? 2) = 0
?7x ? 7 + 8y ?
24 ? 3z + 6 = 0
7x ? 8y + 3z +
25 = 0
which
is the required equation of the plane.
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