Two lines `L_1x=5, y/(3-alpha)=z/(-2)a n dL_2: x=alphay/(-1)=z/(2-alpha)`are coplanar. Then `alpha`can take value (s)a. `1`b. `2`c. `3`d. `4`A. 1,4,5B. 1,2,5C. 3,4,5D. 2,4,5

Two lines `L_1x=5, y/(3-alpha)=z/(-2)a n dL_2: x=alphay/(-1)=z/(2-alph

1.	Two lines `L_1x=5, y/(3-alpha)=z/(-2)a n dL_2: x=alphay/(-1)=z/(2-alpha)`are coplanar. Then `alpha`can take value (s)a. `1`b. `2`c. `3`d. `4`A. 1,4,5B. 1,2,5C. 3,4,5D. 2,4,5
Answer» Correct Answer - A The equations of given lines are: `L_(1):(x-5)/0=y/(3-alpha)=z/(-2),L_(2):(x-alpha)/0=y/(-1)=z/(2-alpha)` These lines are coplanar. Therefore `\|(5-alpha, 0-0,0-0),(0,3-alpha,-2),(0,-1,2-alpha)\|=0` `implies(5-alpha){(3-alpha)(2-alpha)-2}=0` `=((5-alpha)(6-5alpha+alpha^(2)-2)=0` `implies(5-alpha)(alpha^(2)-5alpha+4)=0` `implies(alpha-1)(alpha-4)(alpha-5)=0impliesalpha=1,4,5`

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Two lines `L_1x=5, y/(3-alpha)=z/(-2)a n dL_2: x=alphay/(-1)=z/(2-alpha)`are coplanar. Then `alpha`can take value (s)a. `1`b. `2`c. `3`d. `4`A. 1,4,5B. 1,2,5C. 3,4,5D. 2,4,5

Two lines `L_1x=5, y/(3-alpha)=z/(-2)a n dL_2: x=alphay/(-1)=z/(2-alpha)`are coplanar. Then `alpha`can take value (s)a. `1`b. `2`c. `3`d. `4`A. 1,4,5B. 1,2,5C. 3,4,5D. 2,4,5

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