The answer is: #A=sqrt265#.

There are two ways, the first one ie VERY LONG and complicate, the second one VERY SHORT and easy, but we have to use the vectorial product.

The first one:

First of all, let's check if the shape is really a parallelogram:

#KL=sqrt((x_K-x_L)^2+(y_K-y_L)^2+(x_K-z_L)^2)=#

#=sqrt((1-1)^2+(2-3)^2+(3-6)^2)=sqrt(0+1+9)=sqrt10#.

#MN=sqrt((3-3)^2+(8-7)^2+(6-3)^2)=sqrt(0+1+9)=sqrt10#.

So #KL=MN#

The direction of #KL# is the vector #vecv# such as:

#vecv=(x_K-x_L,y_K-y_L,z_K-z_L)=(0,1,3)#.

The direction of #MN# is the vector #vecw# such as:

#vecw=(x_M-x_N,y_M-y_N,z_M-z_N)=(0,1,3)#.

So #vecv# is parallel to #vecw#.

So, since #KL=MN# and #KL# is parallel to #MN#, the shape is a parallelogram.

The area of a parallelogram is: #A=b*h#.

We can assume that the base #b# is #KL=sqrt10#, but finding the height is more complicated, because it is the distance of the two line #r#, that contains #K and L#, and #s#, that contains #M and N#.

A plane, perpendicular to a line, can be written:

#a(x-x_P)+b(y-y_P)+c(z-z_P)=0#,

where #vecd(a,b,c)# is a whatever vector perpendicular to the plan, and #P# is a whaterver point that lies on the plan.

To find #pi#, that is a plan perpendicular to #r#, we can assume that #vecd=vecv# and #P=K#.

So:

#pi: 0(x-1)+1(y-2)+3(z-3)=0rArry+3z-11=0#.

A line can be written as the system of three equation in parametric form:

#x=x_P+at#

#y=y_P+bt#

#z=z_P+ct#

Where #P# is a whatever point of the line and #vecd(a,b,c)# is a whatever vector, direction of the line.

To find #s#, we can assume that #P=M#, and #vecd=vecw#.

So #s#:

#x=3+0t#

#y=8+1t#

#z=6+3t#

or:

#x=3#

#y=8+t#

#z=6+3t#.

Now, solving the system between #pi# and #s# we can find #Q#, foot of the height conducted from #K# to #s#.

#y+3z-11=0#

#x=3#

#y=8+t#

#z=6+3t#

#8+t+3(6+3t)-11=0rArr10t=-15rArrt=-3/2#.

So, to find the point #Q#, it is necessary to put #t=-3/2# in the equation of #s#.

#x=3#

#y=8-3/2#

#z=6+3(-3/2)#

So:

#x=3#

#y=13/2#

#z=3/2#

Now, to find #h#, we can use the formula of the distance of two points, #K and Q#, just seen before:

#h=sqrt((1-3)^2+(2-13/2)^2+(3-3/2)^2)=sqrt(2^2+(9/2)^2+(3/2)^2)=sqrt(4+81/4+9/4)=sqrt((16+81+9)/4)=sqrt106/2#.

Finally the area is:

#A=sqrt10sqrt106/2=sqrt1060/2=sqrt(4*265)/2=sqrt265#.

The second one.

We can remember that the vectorial product between two vectors is a vector whose lenghts is the area of the parallelogram that has the two vector as two sides.

The vector: #vec(KL)=(0,1,3)#,

the vector #vec(KM)=(2,6,3)#.

And now we have to do: #vec(KL)xxvec(KM)#

We can build the matrix:

first row: #[i,j,k]#,

second row #[0,1,3]#,

third row#[2,6,3]#.

The determinant is the vector: #-15veci+6vecj-2veck#, and his lenghth is: #sqrt(225+36+4)=sqrt265# that is the area requested.

Describe your changes (optional) 200