JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 7)
Let $$(\alpha, \beta, \gamma)$$ be the foot of perpendicular from the point $$(1,2,3)$$ on the line $$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$$. Then $$19(\alpha+\beta+\gamma)$$ is equal to :
99
102
101
100
Explanation
_30th_January_Morning_Shift_en_7_1.png)
Let foot $$P(5 k-3,2 k+1,3 k-4)$$
DR's $$\rightarrow$$ AP : $$5 \mathrm{k}-4,2 \mathrm{k}-1,3 \mathrm{k}-7$$
DR's $$\rightarrow$$ Line: $$5,2,3$$
Condition of perpendicular lines $$(25 k-20)+(4 k-2)+(9 k-21)=0$$
Then $$\mathrm{k}=\frac{43}{38}$$
Then $$19(\alpha+\beta+\gamma)=\mathbf{1 0 1}$$
Comments (0)
