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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 4th April Morning Shift

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: