I have tried like this,

Case I: One 7 and two cards from {1,2,3,4,5,6}

$6C2=15$ ways.

Case II: Two 7's and two cards from {1,2,3,4,5,6}

$6C1=6$ ways.

Case III: Three 7's

$1$ way.

Case IV: Three cards from {1,2,3,4,5,6}

$3$ odds, $3$ evens. To get atleast 1 odd,

$3C1 \times 3C2 +3C2 \times 3C1=18$

Hence, total possible ways is $15+6+1+18=40$ ways.

I have tried like this,

Case I: One 7 and two cards from {1,2,3,4,5,6}

$6C2=15$ ways.

Case II: Two 7's and two cards from {1,2,3,4,5,6}

$6C1=6$ ways.

Case III: Three 7's

$1$ way.

Case IV: Three cards from {1,2,3,4,5,6}

$3$ odds, $3$ evens. To get atleast 1 odd,

$3C3\times3C0[three\ odds]+ 3C1 \times 3C2[one\ odd] +3C2 \times 3C1[two\ odds]=19$

Hence, total possible ways is $15+6+1+19=41$ ways.