Let in a ΔABC, the length of the side AC be 6 , the vertex B be (1, 2, 3) and the vertices A, C lie on the line x-6/3=y-7/2=z-7/-2.

Question

Let in a \(\triangle \mathrm{ABC},\) the length of the side AC be 6 , the vertex B be (1, 2, 3) and the vertices A, C lie on the line \(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}.\) Then the area (in sq. units) of \(\triangle A B C\) is

(1) 42

(2) 21

(3) 56

(4) 17

Answer 1

Correct option is (2) 21

\(Area =\frac{1}{2} \times h(A C)=3 h\)

Where h is minimum distance

\((3 \lambda+6,2 \lambda+7,-2 \lambda+7)\)

\(3(3 \lambda+5)+2(2 \lambda+5)+(-2)(-2 \lambda+4)=0 \)

\(\Rightarrow B D=7=h \)

\(\text { Area }=3(7)=21\)