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Let the image of the point \( P(0,-5,0) \) in the line \( \dfrac{x-1}{2}=\dfrac{y}{1}=\dfrac{z+1}{-2} \) be the point \( R \) and the image of the point \( Q\!\left(0,\dfrac{-1}{2},0\right) \) in the line \( \dfrac{x-1}{-1}=\dfrac{y+9}{4}=\dfrac{z+1}{1} \) be the point \( S \). Then the square of the area of the parallelogram \( PQRS \) is ________.

Let \( P(\alpha, \beta, \gamma) \) be the point on the line \( \frac{x-1}{2} = \frac{y+1}{-3} = z \) at a distance \( 4\sqrt{14} \) from the point \( (1, -1, 0) \) and nearer to the origin. Then the shortest distance, between the lines \( \frac{x - \alpha}{1} = \frac{y - \beta}{2} = \frac{z - \gamma}{3} \) and \( \frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1} \), is equal to:

Let a triangle \( \triangle PQR \) be such that \( P \) and \( Q \) lie on the line \( \frac{x+3}{8} = \frac{y-4}{2} = \frac{x+1}{2} \) and are at a distance of \( 6 \) units from \( R(1, 2, 3) \). If \( (\alpha, \beta, \gamma) \) is the centroid of \( \triangle PQR \), then \( \alpha + \beta + \gamma \) is equal to:

Let the foot of perpendicular from the point \( (\lambda,2,3) \) on the line \( \dfrac{x-4}{1}=\dfrac{y-9}{2}=\dfrac{z-5}{1} \) be the point \( (1,\mu,2) \). Then the distance between the lines \( \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z+4}{6} \) and \( \dfrac{x-\lambda}{2}=\dfrac{y-\mu}{3}=\dfrac{z+5}{6} \) is equal to:

Let the direction cosines of two lines satisfy the equations: \( 4l + m - n = 0 \) and \( 2mn + 10nl + 3lm = 0 \). Then the cosine of the acute angle between these lines is:

If the point of intersection of the lines \( \frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} \) and \( \frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} \) lies on \( xy \)-plane, then the value of \( a+b \) is:

If the distance of the point \( (a, 2, 5) \) from the image of the point \( (1, 2, 7) \) in the line \( \frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2} \) is \( 4 \), then the sum of all possible value of \( a \) is equal to:

Let the line \( L_1 \) be parallel to the vector \( -3\hat{\text{i}} + 2\hat{\text{j}} + 4\hat{\text{k}} \) and pass through the point \( (2,\, 6,\, 7) \), and the line \( L_2 \) be parallel to the vector \( 2\hat{\text{i}} + \hat{\text{j}} + 3\hat{\text{k}} \) and pass through the point \( (4,\, 3,\, 5) \). If the line \( L_3 \) is parallel to the vector \( -3\hat{\text{i}} + 5\hat{\text{j}} + 16\hat{\text{k}} \) and intersects the lines \( L_1 \) and \( L_2 \) at the points \( C \) and \( D \), respectively, then \( |\overline{CD}|^2 \) is equal to:

Let \( (\alpha, \beta, \gamma) \) be the co-ordinates of the foot of the perpendicular drawn from the point \( (5, 4, 2) \) on the line \( \vec{r} = (\hat{-i} + 3\hat{j} + \hat{k}) + \lambda (2\hat{i} + 3\hat{j} - \hat{k}) \). Then the length of the projection of the vector \( \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \) on the vector \( 6\hat{i} + 2\hat{j} + 3\hat{k} \) is:

The square of the distance of the point \( (-2,-8,6) \) from the line \( \dfrac{x-1}{1}=\dfrac{y-1}{2}=\dfrac{z}{-1} \) along the line \( \dfrac{x+5}{1}=\dfrac{y+5}{-1}=\dfrac{z}{2} \) is equal to:

Let \( L \) be the line \( \frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6} \) and let \( S \) be the set of all points \( (a, b, c) \) on \( L \), whose distance from the line \( \frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0} \) along the line \( L \) is \( 7 \). Then \( \sum_{(a,b,c)\in S} (a + b + c) \) is equal to:

If \( \left(2\alpha+1,\alpha^2-3\alpha,\frac{\alpha-1}{2}\right) \) is the image of \( (\alpha,2\alpha,1) \) in the line \( \frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1} \), then the possible value(s) of \( \alpha \) is (are):

Let a line \( L_1 \) pass through the origin and be perpendicular to the lines \( L_2:\ \mathbf{\vec r}=(3+t)\mathbf{\hat i}+(2t-1)\mathbf{\hat j}+(2t+4)\mathbf{\hat k} \) and \( L_3:\ \mathbf{\vec r}=(3+2s)\mathbf{\hat i}+(3+2s)\mathbf{\hat j}+(2+s)\mathbf{\hat k},\ t,s\in \mathbb{R} \). If \( (a,b,c),\ a\in \mathbb{Z} \), is the point on \( L_3 \) at a distance of \( \sqrt{17} \) from the point of intersection of \( L_1 \) and \( L_2 \), then \( (a+b+c)^2 \) is equal to \( \_\_\_\_\_ \).

The sum of all values of \( \alpha \), for which the shortest distance between the lines \( \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} \) and \( \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} \) is \( \sqrt{2} \), is:

Let a line \( L \) passing through the point \( P(1, 1, 1) \) be perpendicular to the lines \( \frac{x-4}{4} = \frac{y-1}{1} = \frac{z-1}{1} \) and \( \frac{x-17}{1} = \frac{y-71}{1} = \frac{z}{0} \). Let the lines \( L \) intersect the \( yz \)-plane at the point \( Q \). Another line parallel to \( L \) and passing through the point \( S(1, 0, -1) \) intersects the \( yz \)-plane at the point \( R \). Then the square of the area of the parallelogram \( PQRS \) is equal to _______.

The vertices \( B \) and \( C \) of a triangle \( ABC \) lie on the line \( \dfrac{x}{1} = \dfrac{1 - y}{-2} = \dfrac{z - 2}{3} \). The coordinates of \( A \) and \( B \) are \( (1, 6, 3) \) and \( (4, 9, \alpha) \) respectively and \( C \) is at a distance of \( 10 \) units from \( B \). The area (in sq. units) of \( \triangle ABC \) is:

If the distances of the point \( (1,2,a) \) from the line \( \dfrac{x-1}{1} = \dfrac{y}{2} = \dfrac{z-1}{1} \) along the lines \( L_{1}: \ \dfrac{x-1}{3} = \dfrac{y-2}{4} = \dfrac{z-a}{b} \) and \( L_{2}: \ \dfrac{x-1}{1} = \dfrac{y-2}{4} = \dfrac{z-a}{c} \) are equal, then \( a+b+c \) is equal to:

Let the point \( A \) be the foot of perpendicular drawn from the point \( P(a,b,0) \) on the line \( \dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-\alpha}{3} \). If the midpoint of the line segment \( PA \) is \( \left(0,\dfrac{3}{4},-\dfrac{1}{4}\right) \), then the value of \( a^2+b^2+\alpha^2 \) is equal to:

Let the lines $L_1:\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k}),\lambda\in R$ and $L_2:\vec{r}=(4\hat{i}+\hat{j})+\mu(5\hat{i}+2\hat{j}+\hat{k}),\mu\in R$, intersect at the point $R$. Let $P$ and $Q$ be the point lying on lines $L_1$ and $L_2$, respectively, such that $\left|\overrightarrow{PR}\right|=\sqrt{29}$ and $\left|\overrightarrow{PQ}\right|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to:

If the image of the point \( P(1, 2, a) \) in the line \( \frac{x-6}{3} = \frac{y-7}{2} = \frac{7 - z}{2} \) is \( Q(5, b, c) \), then \( a^{2} + b^{2} + c^{2} \) is equal to:

Let a line \( L \) be perpendicular to both the lines \( L_{1} : \, \dfrac{x+1}{3} = \dfrac{y+3}{5} = \dfrac{z+5}{7} \) and \( L_{2} : \, \dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-6}{7} \). If \( \theta \) is the acute angle between the lines \( L \) and \( L_{3} : \, \dfrac{x-8/7}{2} = \dfrac{y-4/7}{1} = \dfrac{z}{2} \), then \( \theta \) is equal to:

A line with direction ratios \( 1,-1,2 \) intersects the lines \( \dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z+1}{3} \) and \( \dfrac{x+1}{-1}=\dfrac{y-2}{1}=\dfrac{z}{4} \) at the points \( P \) and \( Q \), respectively. If the length of the line segment \( PQ \) is \( a \), then \( 225a^{2} \) is equal to:

Let a line \( L \) passing through the point \( (1,1,1) \) be perpendicular to both the vectors \( 2\hat{i}+2\hat{j}+\hat{k} \) and \( \hat{i}+2\hat{j}+2\hat{k} \). If \( P(a,b,c) \) is the foot of perpendicular from the origin on the line \( L \), then the value of \( 34(a+b+c) \) is:

Let the image of the point \( P(1,6,a) \) in the line \( L : \, \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-a+1}{b}, \, b>0 \), be \( \left(\dfrac{a}{3},0,a+c\right) \). If \( S(\alpha,\beta,\gamma), \, \alpha>0 \), is the point on \( L \) such that the distance of \( S \) from the foot of perpendicular from the point \( P \) on \( L \) is \( 2\sqrt{14} \), then \( \alpha + \beta + \gamma \) is equal to:

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JEE Main 2026 5th April Evening Shift

If the distance of the point \( (a, 2, 5) \) from the image of the point \( (1, 2, 7) \) in the line \( \frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2} \) is \( 4 \), then the sum of all possible value of \( a \) is equal to: