We have,
x=logabc
x=logbcloga [∵logam=logmloga]
Similarly,
y=logaclogb
z=logablogc
Consider, 1x+1+1y+1+1z+1
⇒1logbcloga+1+1logaclogb+1+1logablogc+1
⇒logalogbc+loga+logblogac+logb+logclogab+logc
⇒logalogabc+logblogabc+logclogabc
⇒loga+logb+logclogabc
⇒logabclogabc
⇒1
Hence, this is the answer.