It is given that X-intercept (a)=2 Y -intercept (b)=3 Z -intercept (c)=4 Therefore, the equation of the plane is,
x
2
+
y
3
+
z
4
=1 6x+4y+3z=12 The given points are A(−1,6,2),B(1,2,3) and C(−2,3,4) Thus, DR's of BC=(−2−1,3−2,4−3) =(−3,1,1) Therefore, the equation of plane passing through A(−1,6,2) and DR′s(−3,1,1) is, a(x−x1)+b(y−y1)+c(z−z1)=0 −3(x+1)+1(y−6)+1(z−2)=0 −3x+y+z−11=0 −3x+y+z=−11 Hence, the angle between the planes is cosθ=