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**arze** Prove that the line of intersection of the two planes

$\displaystyle r=\lambda_1i+2\mu_1j+3\nu_1k$ $\displaystyle \lambda_1+\mu_1+\nu_1=1$

$\displaystyle r=2\lambda_2i+\mu_1j+2\nu_1k$ $\displaystyle \lambda_2+\mu_2+\nu_2=1$

Too many equality signs...

can be written in terms of a single parameter *t* as

$\displaystyle 6r=(3+t)i+4tj+9(1-t)k$

I managed to get the vector equations into this form:

$\displaystyle r.(6i+3j+2k)=6$

and $\displaystyle r.(i+2j+k)=4$

I calculated the direction vector of the line to be (-*i*-4*j*+9*k*)

I don't know how else to continue.

Thanks!